RD Sharma Solutions Class 10 Trigonometric Identities Exercise 6.1

Exercise 6.1

Prove the following trigonometric identities

Q1: (1-Cos² A) Cosec² A = 1

Ans: (1-Cos² A) Cosec² A = Sin² A Cosec² A

= (Sin A Cosec A)²

= (Sin A x (1/Sin A))²

= (1)² = 1

Q2: (1 + Cot² A) Sin² A = 1

Ans: We know, Cosec² A –Cot ² A = 1

So,

(1 + Cot² A) Sin² A = Cosec² A Sin² A

= (Cosec A Sin A)²

= ((1/Sin A) x Sin A )²

= (1)² = 1

Q3:  \(tan^{2}\theta\;cos^{2}\theta\)= \(1-cos^{2}\theta\)

A3: We know ,

\(sin^{2}\theta + cos^{2}\theta=1\)

So,

\(tan^{2}\theta\;cos^{2}\theta\) = \((tan\theta \times cos\theta)^{2}\)

\(= (\frac{sin\theta}{cos\theta}\times cos\theta)^{2}\)

\(= (sin\theta)^{2}\)

\(= sin^{2}\theta\)

\(1-cos^{2}\theta\)

Q4: \(cosec\theta\sqrt{1-cos^{2}\theta}=1\)

A4: We know ,

\(sin^{2}\theta + cos^{2}\theta=1\)

So,

\(cosec\theta\sqrt{1-cos^{2}\theta}=cosec\theta\sqrt{sin^{2}\theta}\)

= \(=cosec\theta \;sin\theta\)

= \(=\frac{1}{sin\theta}\;sin\theta\)

= 1

Q5 : \((sec^{2}\theta-1)(cosec^{2}\theta-1)=1\)

A5: We know that,

\((sec^{2}\theta-tan^{2}\theta)= 1\)

\((cosec^{2}\theta-cot^{2}\theta)= 1\)

So,

\((sec^{2}\theta-1)(cosec^{2}\theta-1)= tan^{2}\theta\times cot^{2}\theta\)

\(= (tan\theta\times cot\theta)^{2}\)

\(= (tan\theta\times \frac{1}{tan\theta})^{2}\)

= 1² = 1

Q6:  \(tan\theta+ \frac{1}{tan\theta} =sec\theta\;cosec\theta\)

A6: We know that,

\((sec^{2}\theta-tan^{2}\theta)= 1\)

So,

\(tan\theta+ \frac{1}{tan\theta} =\frac{tan^{2}\theta+1}{tan\theta}\)

\(=\frac{sec^{2}\theta}{tan\theta}\)

\(=sec\theta\frac{sec\theta}{tan\theta}\)

\(=sec\theta\frac{\frac{1}{cos\theta}}{\frac{sin\theta}{cos\theta}}\)

\(=sec\theta\frac{1}{sin\theta}\)

\(=sec\theta cosec\theta\)

Q7: \(\frac{cos\theta}{1-sin\theta}=\frac{1+sin\theta}{cos\theta}\)

A7:  We know ,

\(sin^{2}\theta + cos^{2}\theta=1\)

So, Multiplying both numerator and denominator by  \({(1+sin\theta)}\), we have

\(\frac{cos\theta}{1-sin\theta}=\frac{cos\theta(1+sin\theta)}{(1-sin\theta)(1+sin\theta)}\)

\(=\frac{cos\theta(1+sin\theta)}{(1-sin^{2}\theta)}\)

\(=\frac{cos\theta(1+sin\theta)}{cos^{2}\theta}\)

\(=\frac{(1+sin\theta)}{cos\theta}\)

Q8: \(\frac{cos\theta}{1+sin\theta}=\frac{1-sin\theta}{cos\theta}\)

A8: We know ,

\(sin^{2}\theta + cos^{2}\theta=1\)

Multiplying both numerator and denominator by \({(1-sin\theta)}\), we have

\(\frac{cos\theta}{1+sin\theta}=\frac{cos\theta(1-sin\theta)}{(1+sin\theta)(1-sin\theta)}\)

\(=\frac{cos\theta(1-sin\theta)}{(1-sin^{2}\theta)}\)

\(=\frac{cos\theta(1-sin\theta)}{(cos^{2}\theta)}\)

\(=\frac{(1-sin\theta)}{cos\theta}\)

\(=\frac{(1-sin\theta)}{cos\theta}\)

Q 9:cos²A +\(\frac{1}{1+cot^{2}A}\) = 1

A9: We know that,

Sin²A + cos²A = 1

cosec²A – cot²A = 1

So, \(cos^{2}A+\frac{1}{1+cot^{2}A} = cos^{2}A + \frac{1}{cosec^{2}A}\)

\(= cos^{2}A + (\frac{1}{cosecA})^{2}\)

\(= cos^{2}A + sinA^{2}\)

= 1

Q10: \(sinA^{2}+\frac{1}{1+tan^{2}A}=1\)

A10:     We know,

Sin²A + cos²A = 1

sec²A – tan²A = 1

                        So,

\(sinA^{2}+\frac{1}{1+tan^{2}A}=sinA^{2}+\frac{1}{sec^{2}A}\)

\(=sinA^{2}+(\frac{1}{secA})^{2}\)

\(=sinA^{2}+cos^{2}A\)

= 1

Q11: \(\sqrt{\frac{1-cos\theta }{1+cos\theta}}=cosec\theta-cot\theta\)

A11:  We know ,

\(sin^{2}\theta + cos^{2}\theta=1\)

Multiplying both numerator and denominator by \({(1-cos\theta)}\), we have

\(\sqrt{\frac{1-cos\theta }{1+cos\theta}}=\sqrt{\frac{(1-cos\theta)(1-cos\theta) }{(1+cos\theta)(1-cos\theta)}}\)

\(=\sqrt{\frac{(1-cos\theta)^{2}}{1-cos^{2}\theta}}\)

\(=\sqrt{\frac{(1-cos\theta)^{2}}{sin^{2}\theta}}\)

\(={\frac{(1-cos\theta)}{sin\theta}}\)

\(=\frac{1}{sin\theta}-\frac{cos\theta}{sin\theta}\)

\(=\frac{1}{sin\theta}-\frac{cos\theta}{sin\theta}[/latex\]

Q12: \(\frac{1-cos\theta }{sin\theta}=\frac{sin\theta}{1+cos\theta}\)

Q12: \(\frac{1-cos\theta }{sin\theta}=\frac{sin\theta}{1+cos\theta}\]”>

A12: We know ,

\(sin^{2}\theta + cos^{2}\theta=1\)

Multiplying both numerator and denominator by \({(1+cos\theta)}\), we have

\(=\frac{(1-cos^{2}\theta)}{(1+cos\theta)(sin\theta)}\)

\(=\frac{(sin^{2}\theta)}{(1+cos\theta)(sin\theta)}\)

\(=\frac{(sin\theta)}{(1+cos\theta)}\)

Q13. \(\frac{sin\theta}{1-cos\theta}\) = cosec\(\theta\) + cot\(\theta\)

Ans:

Given, L.H.S = \(\frac{sin\theta}{1-cos\theta}\)

Rationalize both nr and dr with 1+cos \(\theta\)

= \(\frac{sin\theta}{1-cos\theta}\) * \(\frac{1+cos\theta}{1+cos\theta}\)

We know that, (a-b)(a+b) = a² – b²

=> \(\frac{sin\theta(1+cos\theta)}{1-cos^{2}\theta}\)

Here, (1-cos² \(\theta\)) = sin² \(\theta\)

=> \(\frac{sin\theta\;+\;(sin\theta*cos\theta)}{sin^{2}\theta}\)

=> \(\frac{sin\theta }{sin^{2}\theta}\) + \(\frac{sin\theta*cos\theta}{sin^{2}\theta}\)

=> \(\frac{1}{sin\theta}\) + \(\frac{cos\theta}{sin\theta}\)

=> cosec \(\theta\) + cot \(\theta\)

Hence, L.H.S = R.H.S

Q14. \(\frac{1\;-\;sin\theta}{1\;+\;sin\theta}\) = \((sec\theta\;-\;tan\theta)^{2}\)

Ans:

Given, L.H.S = \(\frac{1\;-\;sin\theta}{1\;+\;sin\theta}\)

Rationalize with nr and dr with 1 – sin \(\theta\)

=> \(\frac{1\;-\;sin\theta}{1\;+\;sin\theta}\) * \(\frac{1\;-\;sin\theta}{1\;-\;sin\theta}\)

Here, (1-sin \(\theta\))(1+sin \(\theta\)) = cos² \(\theta\)

=> \(\frac{(1\;-\;sin\theta)^{2}}{cos^{2}\theta}\)

=> \((\frac{1\;-\;sin\theta}{cos\theta})^{2}\)

=> \((\frac{1}{cos\theta}\;-\;\frac{sin\theta}{cos\theta})^{2}\)

=> \((sec\theta\;-\;tan\theta)^{2}\)

Hence,  L.H.S = R.H.S

Q15. \(\frac{(1\;+\;cot^{2}\theta)tan\theta}{sec^{2}\theta}\) = cot\theta

Ans:

Given, L.H.S = \(\frac{(1\;+\;cot^{2}\theta)tan\theta}{sec^{2}\theta}\)

Here, 1 + cot² \(\theta\) =cosec² \(\theta\)

=> \(\frac{cosec^{2}\theta*tan\theta}{sec^{2}\theta}\)

=> \(\frac{1}{sin^{2}\theta}\) * \(\frac{cos^{2}\theta}{1}\) * \(\frac{sin\theta}{cos\theta}\)

=> \(\frac{cos\theta}{sin\theta}\)

=> cot \(\theta\)

Hence,  L.H.S = R.H.S

Q16. \(tan^{2}\theta\;-\;sin^{2}\theta\) = \(tan^{2}\theta\;*\;sin^{2}\theta\)

Ans:

Given, L.H.S = \(tan^{2}\theta\;-\;sin^{2}\theta\)

Here, tan² \(\theta\) = \(\frac{sin^{2}\theta}{cos^{2}\theta}\)

=> \(\frac{sin^{2}\theta}{cos^{2}\theta}\) – \(sin^{2}\theta\)

=> \(sin^{2}\theta\)[ \(\frac{1}{cos^{2}\theta}\) – 1]

=> \(sin^{2}\theta\)[ \(\frac{1\;-\;cos^{2}\theta}{cos^{2}\theta}\)]

=> \(\frac{sin^{2}\theta}{cos^{2}\theta}\) * \(sin^{2}\theta\)

=> \(tan^{2}\theta\;*\;sin^{2}\theta\)

Hence, L.H.S = R.H.S

Q17. (cosec \(\theta\) + sin \(\theta\))(cosec \(\theta\) – sin \(\theta\)) = \(cot^{2}\theta\;+\;cos^{2}\theta\)

Ans:

Given, L.H.S = (cosec \(\theta\) + sin \(\theta\))(cosec \(\theta\) – sin \(\theta\))

Here, (a + b)(a – b) = a² – b²

cosec² \(\theta\) can be written as 1 + cot² \(\theta\) and sin² \(\theta\) can be written as 1 – cos² \(\theta\)

=>  1 + cot² \(\theta\) – (1 – cos² \(\theta\))

=>  1 + cot² \(\theta\) – 1 + cos² \(\theta\)

=> cot² \(\theta\) + cos² \(\theta\)

Hence, L.H.S = R.H.S

Q18. (\(sec\theta\) +cos \(\theta\))(sec \(\theta\) – cos \(\theta\)) = \(tan^{2}\theta\;+\;sin^{2}\theta\)

Ans:

Given, L.H.S = (sec \(\theta\) + cos \(\theta\))(sec \(\theta\) – cos \(\theta\))

Here, (a + b)(a – b) = a² – b²

sec²\(\theta\) can be written as 1 + tan² \(\theta\) and cos² \(\theta\) can be written as 1 – sin² \(\theta\)

=> 1 + tan² \(\theta\) – (1 – sin² \(\theta\))

=> 1 + tan² \(\theta\) – 1 + sin² \(\theta\)

=>  tan²  \(\theta\) + sin² \(\theta\)

Hence, L.H.S = R.H.S

Q19. secA(1 – sinA)(secA + tanA) = 1

Ans:

Given, L.H.S = secA(1 – sinA)(secA + tanA)

Here, secA = \(\frac{1}{cosA}\) and tanA = \(\frac{sinA}{cosA}\)

=> \(\frac{1}{cosA}\) * (1 – sinA) * \(\frac{1 + sinA}{cosA}\)

=> \(\frac{cos²A}{cos²A}\)

=> 1

Hence, L.H.S = R.H.S

Q20. (cosecA – sinA)(secA – cosA)(tanA + cotA) = 1

Ans:

Given, L.H.S = (cosecA – sinA)(secA – cosA)(tanA + cotA)

Here, cosecA = \(\frac{1}{sinA}\), secA = \(\frac{1}{cosA}\), tanA = \(\frac{sinA}{cosA}\), cotA = \(\frac{cosA}{sinA}\)

Substitute the above values in L.H.S

=> (\(\frac{1}{sinA}\) – sinA)( \(\frac{1}{cosA}\) – cosA)( \(\frac{sinA}{cosA}\) + \(\frac{cosA}{sinA}\))

=> (\(\frac{1\;-sin^{2}A}{sinA}\)) * (\(\frac{1\;-cos^{2}A}{cosA}\)) * (

\(\frac{sin^{2}A\;+\;cos^{2}A}{sinA\;*\;cosA}\)

Here, (\(\frac{1\;-sin^{2}A}) = cos^{2}A, (\(\frac{1\;-cos^{2}A}) = sin^{2}A, sin^{2}A + cos^{2}A = 1

=> \(\frac{sin^{2}A\;*\;cos^{2}A\;*\;1}{sin^{2}A\;*\;cos^{2}A}\)\(\frac{1\;-cos^{2}A}) = sin^{2}A, sin^{2}A + cos^{2}A = 1

=> \(\frac{sin^{2}A\;*\;cos^{2}A\;*\;1}{sin^{2}A\;*\;cos^{2}A}\]”>

=> 1

Hence,  L.H.S = R.H.S

Q21. (1 + \(tan^{2}\theta\))(1 – sin \(\theta\))(1 + sin \(\theta\)) = 1

Ans:

Given, L.H.S = (1 + \(tan^{2}\theta\))(1 – sin\theta)(1 + sin\theta)

We know that,

Sin² \(\theta\) + cos² \(\theta\) = 1

And sec² \(\theta\) – tan² \(\theta\) = 1

So,

(1 + \(tan^{2}\theta\))(1 – sin \(\theta\))(1 + sin \(\theta\)) = (1 + \(tan^{2}\theta\)){(1 – sin \(\theta\))(1 + sin \(\theta\))}

= (1 + \(tan^{2}\theta\))( (1 – \(sin^{2}\theta\))

= \(sec^{2}\theta\;*\;tan^{2}\theta\)

= \((\frac{1}{cos^{2}\theta})\;*\;cos^{2}\theta\)

= 1

hence, L.H.S = R.H.S

Q22. \((sin^{2}A\;*\;cot^{2}A)\;+\;(cos^{2}A;*\;tan^{2}A)\) = 1

Ans:

Given, L.H.S = \((sin^{2}A\;*\;cot^{2}A)\;+\;(cos^{2}A\;*\;tan^{2}A)\)

Here, \((sin^{2}A\;+\;cos^{2}A) \)= 1

So,

\((sin^{2}A\;*\;cot^{2}A)\;+\;(cos^{2}A\;*\;tan^{2}A)\)

So,

\((sin^{2}\A\;*\;cot^{2}A)\;+\;(cos^{2}A\;*\;tan^{2}A)[\latex]

= sin²A(\(\frac{cos^{2}A}{sin^{2}A}\)) + cos²A(\(\frac{sin^{2}A}{cos^{2}A}\)

= cos²A + sin²A

= 1

Hence , L.H.S = R.H.S

Q23:

1. cot \(\theta\) – tan \(\theta\) = \(\frac{2cos^{2}\theta\;-\;1}{sin\theta*cos\theta}\)

Ans:

Give, L.H.S = cot \(\theta\) – tan \(\theta\)

Here, Sin² \(\theta\) + cos² \(\theta\) = 1

So,

=> cot \(\theta\) – tan \(\theta\) = \(\frac{cos\theta}{sin\theta}\) – \(\frac{sin\theta}{cos\theta}\)

= \(\frac{cos^{2}\theta\;-\;sin^{2}\theta}{sin\theta\;*\;cos\theta }\)

= \(\frac{cos^{2}\theta\;-\;(1\;-\;cos^{2}\theta)}{sin\theta\;*\;cos\theta }\)

= \(\frac{cos^{2}\theta\;-\;1\;-\;cos^{2}\theta}{sin\theta\;*\;cos\theta }\)

= \((\frac{2cos^{2}\theta\;-\;1}{sin\theta*cos\theta})\)

Hence, L.H.S = R.H.S

1. \(tan\theta\;-\;cot\theta\;=\;\) \((\frac{2sin^{2}\theta\;-\;1}{sin\theta*cos\theta})\)

Sol:

Given, L.H.S = \(tan\theta\;-\;cot\theta

We know that,

Sin² \(\theta\)

We know that,

Sin² \(\theta\]”>

+ cos² \(\theta\) = 1

\(tan\theta\;-\;cot\theta = \(\frac{sin\theta }{cos\theta }\)\(\frac{sin\theta }{cos\theta }\]”> – \(\frac{cos\theta }{sin\theta }\)

= \(\frac{sin^{2}\theta\;-\;cos^{2}\theta }{sin\theta cos\theta }\)

= \(\frac{sin^{2}\theta\;-\;(1\;-\;sin^{2}\theta) }{sin\theta cos\theta }\)

= \(\frac{sin^{2}\theta\;-\;1\;+\;sin^{2}\theta }{sin\theta cos\theta }\)

= \((\frac{2sin^{2}\theta\;-\;1}{sin\theta*cos\theta})\)

Hence,  L.H.S = R.H.S

Q24. \(\frac{cos^{2}\theta }{sin\theta }\;-\;cosec\theta\;+\;sin\theta\) = 0

Ans:

Given, L.H.S \(\frac{cos^{2}\theta }{sin\theta }\;-\;cosec\theta\;+\;sin\theta\)

We know that,

Sin² \(\theta\) + cos² \(\theta\) = 1

So,

\(\frac{cos^{2}\theta }{sin\theta }\;-\;cosec\theta\;+\;sin\theta\) = (\(\frac{cos^{2}\theta }{sin\theta }\;-\;cosec\theta)\;+\;sin\theta\)

= \((\frac{cos^{2}\theta }{sin\theta }\;-\;\frac{1}{sin\theta })\;+\;sin\theta\)

= \((\frac{cos^{2}\theta\;-\;1 }{sin\theta })\;+\;sin\theta\)

= \((\frac{-sin^{2}\theta }{sin\theta })\;+\;sin\theta\)

= \(-sin\theta \;+\;sin\theta\)

= 0

Hence, L.H.S = R.H.S

Q 25 . \(\frac{1}{1 + sin\;A}\) + \(\frac{1}{1 – sin\;A}\) = 2 sec² A

Ans:

LHS = \(\frac{1}{1 + sin\;A}\) + \(\frac{1}{1 – sin\;A}\)

\(\frac{\left ( 1-sinA \right )+\left ( 1+sinA \right )}{\left ( 1+sinA \right )\left ( 1-sinA \right )}\)

\(\frac{1-sinA+1+sinA}{1-sin^{2}A}\)

\(\Rightarrow \frac{2}{1-sin^{2}\;A}\)                               [Since , (1 + sin A)(1 – sin A) = \(1-sin^{2} A\)]

\(\Rightarrow \frac{2}{cos^{2}A}\)                                    [Since , 1-sin² A = cos2 A]

\(\Rightarrow 2sec^{2}A\)

∴ LHS = RHS   Hence proved

Q 26 . \(\frac{1+sin \theta }{cos \theta}+\frac{cos \theta}{1 + sin \theta} = 2sec \theta\)

Ans:

LHS = \(\frac{1+sin \theta }{cos \theta}+\frac{cos \theta}{1 + sin \theta}\)

= \(\frac{\left (1+sin \theta \right )^{2}+cos^{2}\theta }{cos \theta\left ( 1 + sin \theta \right )}\)

= \(\frac{1+sin^{2}\theta +2sin \theta +cos^{2}\theta }{cos \theta\left ( 1 + sin \theta \right )}\)

\(\Rightarrow \frac{2\left ( 1+sin \theta \right ) }{cos \theta\left ( 1 + sin \theta \right )}\) = 2 \(sec\theta\)

∴ LHS = RHS   Hence proved

Q 27 . \(\frac{\left (1+sin\theta \right )^{2}+\left (1-sin\theta \right )^{2} }{2cos^{2}\theta }=\frac{1+sin^{2}\theta }{1-sin^{2}\theta }\)

Ans:

We know that \(sin^{2}\theta +cos^{2}\theta =1\)

So,

LHS = \(\frac{\left ( 1+sin\theta \right )^{2}+\left ( 1-sin\theta \right )^{2}}{2cos^{2}\theta }\\=\frac{\left ( 1+2sin\theta+sin^{2} \theta \right )+\left ( 1-2sin\theta+sin^{2} \theta \right )}{2cos^{2}\theta}\\=\frac{ 1+2sin\theta+sin^{2} \theta +1-2sin\theta+sin^{2} \theta }{2cos^{2}\theta}\\=\frac{2+2sin^{2}\theta }{2cos^{2}\theta }\\=\frac{2\left ( 1+sin^{2}\theta \right )}{2\left ( 1-sin^{2}\theta \right )}\\=\frac{\left ( 1+sin^{2}\theta \right )}{\left ( 1-sin^{2}\theta \right )}\)

∴ LHS = RHS   Hence proved

Q 28 . \(\frac{1+tan^{2}\theta }{1+cot^{2}\theta } = \left [ \frac{1-tan\theta }{cot\theta } \right ]^{2}-tan^{2}\theta\)

Ans :

LHS = \(\frac{1+tan^{2}\theta }{1+cot^{2}\theta }\)

= \(\frac{sec^{2}\theta }{cosec^{2}\theta }\)                                                        [Since , \(tan^{2}\theta\) + 1 = \(sec^{2}\theta\) , 1 + \(cot^{2}\theta\) = \(cosec^{2}\theta\)]

= \(\frac{1}{cos^{2}\theta \cdot 1}sin^{2}\theta\)

= \(tan^{2}\theta\)

∴ LHS = RHS   Hence proved

Q 29 . \(\frac{1+sec\theta} {sec\theta}\) = \(\frac{sin^{2}\theta }{1-cos\theta }\)

Ans :

LHS =  \(\frac{1+sec\theta} {sec\theta}\)

= \(\frac{1+\frac{1}{cos\theta }}{\frac{1}{cos\theta }}\)

= \(\frac{cos\theta +1}{cos\theta }\cdot cos\theta\)

= \(1 + cos\theta\)

RHS = \(\frac{sin^{2}\theta }{1-cos\theta }\)

= \(\frac{1 – cos^{2}\theta }{1-cos\theta }\)

= \(\frac{\left ( 1-cos\theta \right )\left ( 1+cos\theta \right )}{1-cos\theta }\)

= \(1 + cos\theta\)

∴ LHS = RHS   Hence proved

Q 30 . \(\frac{tan\theta }{1-cot\theta }+\frac{cot\theta }{1-tan\theta}\) = \(1 + tan\theta +cot\theta\)

Ans:

LHS = \(\frac{tan\theta }{1-\frac{1}{tan\theta }}+\frac{cot\theta }{1-tan\theta }\)

= \(\frac{tan^{2}\theta }{tan\theta -1}+\frac{cot\theta }{1-tan\theta }\)

= \(\frac{1 }{1-tan\theta }\left [ \frac{1}{tan\theta } -tan^{2}\theta \right ]\)

= \(\frac{1 }{1-tan\theta }\left [ \frac{1-tan^{3}\theta}{tan\theta } \right ]\)

= \(\frac{1 }{1-tan\theta }\frac{\left ( 1-tan\theta \right )\left ( 1+tan\theta +tan^{2}\theta \right )}{tan\theta}\)                                        [Since , \(a^{3}-b^{3}=\left ( a-b \right )\left ( a^{2}+ab+b^{2} \right )\)]

= \(\frac{ 1+tan\theta +tan^{2}\theta }{tan\theta}\)

= \(\frac{1}{tan\theta }+\frac{tan\theta}{tan\theta}+\frac{tan^{2}\theta}{tan\theta}\)

=  \(1 + tan\theta +cot\theta\)

∴ LHS = RHS   Hence proved

Q 31 . \(sec^{6}\theta = tan^{6}\theta+3 tan^{2} \theta sec^{2}\theta +1\)

Ans :

We know that \(sec^{2}\theta-tan^{2}\theta=1\)

Cubing  both sides

\(\left (sec^{2}\theta -tan^{2}\theta \right )^{3} = 1\)

\(sec^{6}\theta -tan^{6}\theta – 3 sec^{2}\theta tan^{2}\theta \left ( sec^{2}\theta -tan^{2}\theta \right ) = 1\)                                              [Since , \(a^{3}-b^{3}=\left ( a-b \right )\left ( a^{2}+ab+b^{2} \right )\)]

\(sec^{6}\theta -tan^{6}\theta – 3 sec^{2}\theta tan^{2}\theta = 1\)

\(\Rightarrow sec^{6}\theta = tan^{6}\theta+3 sec^{2}\theta tan^{2}\theta +1\)

Hence proved.

Q 32 . \(cosec^{6}\theta = cot^{6}\theta+3 cot^{2} \theta cosec^{2}\theta +1\)

Ans :

We know that \(cosec^{2}\theta –  cot^{2}\theta = 1\)

Cubing  both sides

\(\left (cosec^{2}\theta -cot^{2}\theta \right )^{3} = 1\)

\(cosec^{6}\theta -cot^{6}\theta – 3 cosec^{2}\theta cot^{2}\theta \left ( cosec^{2}\theta -cot^{2}\theta \right ) = 1\)                                           [Since , \(a^{3}-b^{3}=\left ( a-b \right )\left ( a^{2}+ab+b^{2} \right )\)]

\(cosec^{6}\theta -cot^{6}\theta – 3 cosec^{2}\theta cot^{2}\theta = 1\)

\(\Rightarrow cosec^{6}\theta = cot^{6}\theta+3 cosec^{2}\theta cot^{2}\theta +1\)

Hence proved.

Q 33 . \(\frac{\left ( 1+tan^{2}\theta \right )cot\theta }{cosec^{2}\theta } = tan\theta\)

Ans :

We know that \(sec^{2}\theta -tan^{2}\theta = 1\)

Therefore , \(sec^{2}\theta = 1+tan^{2}\theta\)

LHS = \(\frac{sec^{2}\theta \cdot cot\theta }{cosec^{2}\theta }\)

= \(\frac{1\cdot sin^{2}\theta }{cos^{2}\theta }\cdot \frac{cos\theta }{sin\theta }\)                   \(\left [ \;because\; sec\theta = \frac{1}{cos\theta } , cosec\theta = \frac{1}{sin\theta }, cot \theta = \frac{cos \theta }{sin\theta }\right ]\)

\(\Rightarrow \frac{sin\theta }{cos\theta }= tan\theta\)

∴ LHS = RHS   Hence proved

Q 34 . \(\frac{ 1+cosA}{sin ^{2} A}\) = \(\frac{1}{1 – cosA}\)

Ans:

We know that  \(sin^{2} A+cos^{2}A\) = 1

\(sin^{2}A=1-cos^{2}A\)

\(\Rightarrow sin^{2}A=\left (1-cosA \right )\left ( 1+cosA \right )\)

\(\Rightarrow\;LHS = \frac{\left ( 1+cosA \right )}{\left ( 1-cosA \right )\left ( 1+cosA \right )}\)

= \(\Rightarrow\;LHS = \frac{1}{\left ( 1-cosA \right )}\)

∴ LHS = RHS   Hence proved

Q 35 . \(\frac{sec A-tanA}{sec A+tanA}=\frac{cos^{2}A}{\left ( 1+sinA \right )^{2}}\)

Ans:

LHS = \(\frac{sec A-tanA}{sec A+tanA}\)

Rationalizing the denominator by multiplying and dividing with sec A + tan A , we get

\(\frac{sec A-tanA}{sec A+tanA}\times \frac{sec A+tanA}{sec A+tanA}\)

= \(\frac{sec^{2} A-tan^{2}A}{\left (sec A+tanA \right )^{2}}\)

= \(\frac{1}{\left (sec A+tanA \right )^{2}}\)

= \(\frac{1}{\left (sec ^{2}A+tan^{2}A +2sec A tan A \right )}\)

= \(\frac{1}{\left (\frac{1}{cos ^{2}A}+\frac{sin ^{2}A}{cos ^{2}A} +\frac{2sinA}{cosA} \right )}\)

\(\Rightarrow \frac{cos ^{2}A}{1 +sin^{2}A+2 sin A}\)

= \(\frac{cos ^{2}A}{\left (1 +sinA \right )^{2}}\)

∴ LHS = RHS   Hence proved

Q 36 . \(\frac{1+cosA}{sinA}\) = \(\frac{sinA}{1-cosA}\)

Ans:

LHS = \(\frac{1+cosA}{sinA}\)

Multiply both numerator and denominator with (1 – cos A) we get ,

\(\frac{\left (1+cosA \right )\left ( 1-cosA \right )}{sin A\left (1-cosA \right )}\)

= \(\frac{1-cos^{2}A }{sin A\left (1-cosA \right )}\)

= \(\frac{sin^{2}A }{sin A\left (1-cosA \right )}\)

= \(\frac{sinA }{1-cosA }\)

∴ LHS = RHS   Hence proved

37.

 (i) \(\sqrt{\frac{1+sin A}{1- sin A}}\) = sec A + tan A

Ans:

To prove,

\(\sqrt{\frac{1+sin A}{1- sin A}}\) = sec A + tan A

Considering left hand side (LHS),

Rationalize the numerator and denominator with \(\sqrt{1+ sin A}\)

• \(\sqrt{\frac{(1+sin A)(1+ sin A)}{(1-sin A)(1+ sin A}}\) = \(\sqrt{\frac{(1+ sin A)^{2}}{1-sin^{2}A}}\)

= \(\sqrt{\frac{(1+ sin A)^{2}}{cos^{2}A}}\)

= \(\frac{(1+ sin A)}{cosA}\)

= \(\frac{1}{cosA}+\frac{sin A}{cos A}\)

= sec A + tan A

Therefore, LHS = RHS

Hence proved

(ii) \(\sqrt{\frac{(1-cos A)}{(1+cos A)}}+\sqrt{\frac{(1+cos A)}{(1-cos A)}}\) = 2cosec A

Ans:

To prove,

\(\sqrt{\frac{(1-cos A)}{(1+cos A)}}+\sqrt{\frac{(1+cos A)}{(1-cos A)}}\) = 2cosec A

Considering left hand side (LHS),

Rationalize the numerator and denominator.

= \(\sqrt{\frac{(1-cos A)(1-cos A)}{(1+cos A)(1-cos A)}}+\sqrt{\frac{(1+cos A)(1+cos A)}{(1-cos A)(1+cos A)}}\)

= \(\sqrt{\frac{(1-cos A)^{2}}{(1-cos^{2} A)}}+\sqrt{\frac{(1+cos A)^{2}}{(1-cos^{2} A)}}\)

= \(\sqrt{\frac{(1-cos A)^{2}}{(sin^{2} A)}}+\sqrt{\frac{(1+cos A)^{2}}{(sin^{2} A)}}\)

= \(\frac{(1-cos A)}{(sin A)}+\frac{(1+cos A)}{(sin A)}\)

= \(\frac{(1-cos A+1+cos A)}{(sin A)}\)

= \(\frac{(2)}{(sin A)}\)

= 2cosec A

Therefore, LHS = RHS

Hence proved

38. Prove that:

(i) \(\sqrt{\frac{(sec \Theta- 1)}{(sec \Theta+1)}}+\sqrt{\frac{(sec \Theta+ 1)}{(sec \Theta-1)}}\) = 2cosec \(\Theta\)

Ans:

To prove,

= \(\sqrt{\frac{(sec \Theta- 1)}{(sec \Theta+1)}}+\sqrt{\frac{(sec \Theta+ 1)}{(sec \Theta-1)}}\) = 2cosec \(\Theta\)

Considering left hand side (LHS),

Rationalize the numerator and denominator.

= \(\sqrt{\frac{(sec \Theta- 1)(sec \Theta- 1)}{(sec \Theta+1)(sec \Theta-1)}}+\sqrt{\frac{(sec \Theta+ 1)(sec \Theta+1)}{(sec \Theta-1)(sec \Theta+1)}}\)

= \(\sqrt{\frac{(sec \Theta- 1)^{2}}{(sec^{2} \Theta-1)}}+\sqrt{\frac{(sec \Theta+ 1)^{2}}{(sec^{2} \Theta-1)}}\)

= \(\sqrt{\frac{(sec \Theta- 1)^{2}}{tan^{2}\Theta }}+\sqrt{\frac{(sec \Theta+ 1)^{2}}{tan^{2}\Theta}}\)

= \(\frac{(sec \Theta- 1)}{tan\Theta }+\frac{(sec \Theta+ 1)}{tan\Theta}\)

= \(\frac{(sec \Theta- 1+sec \Theta+ 1)}{tan\Theta }\)

= \(\frac{(2cos \Theta)}{cos\Theta sin\Theta }\)

= \(\frac{2}{sin\Theta }\)

= 2cosec \(\Theta\)

Therefore, LHS = RHS

Hence proved

(ii) \(\sqrt{\frac{( 1+sin\Theta)}{( 1-sin\Theta)}}+\sqrt{\frac{( 1-sin\Theta)}{( 1+sin\Theta)}}\) = 2sec \(\Theta\)

Ans:

To prove,

= \(\sqrt{\frac{( 1+sin\Theta)}{( 1-sin\Theta)}}+\sqrt{\frac{( 1-sin\Theta)}{( 1+sin\Theta)}}\) = 2sec \(\Theta\)

Considering left hand side (LHS),

Rationalize the numerator and denominator.

= \(\sqrt{\frac{( 1+sin\Theta)( 1+sin\Theta)}{( 1-sin\Theta)( 1+sin\Theta)}}+\sqrt{\frac{( 1-sin\Theta)( 1-sin\Theta)}{( 1+sin\Theta)( 1-sin\Theta)}}\)

= \(\sqrt{\frac{( 1+sin\Theta)^{2}}{( 1-sin^{2}\Theta)}}+\sqrt{\frac{( 1-sin\Theta)^{2}}{( 1-sin^{2}\Theta)}}\)

= \(\sqrt{\frac{( 1+sin\Theta)^{2}}{( cos^{2}\Theta)}}+\sqrt{\frac{( 1-sin\Theta)^{2}}{( cos^{2}\Theta)}}\)

= \(\frac{( 1+sin\Theta)}{( cos\Theta)}+\frac{( 1-sin\Theta)}{( cos\Theta)}\)

= \(\frac{( 1+sin\Theta+ 1-sin\Theta)}{( cos\Theta)}\)

= \(\frac{(2)}{( cos\Theta)}\)

= \(2sec\Theta\)

Therefore, LHS = RHS

Hence proved

(iii) \( \sqrt{\frac{(1+cos \Theta)}{(1-cos \Theta)}}\) \sqrt{\frac{(1-cos \Theta)}{(1+cos \Theta)}} = \(2cosec\Theta\)

Ans:

To prove,

\(\sqrt{\frac{(1-cos \Theta)}{(1+cos \Theta)}}+\sqrt{\frac{(1+cos \Theta)}{(1-cos \Theta)}}\) = 2cosec \Theta

Considering left hand side (LHS),

Rationalize the numerator and denominator.

= \(\sqrt{\frac{(1-cos \Theta)(1-cos \Theta)}{(1+cos \Theta)(1-cos \Theta)}}+\sqrt{\frac{(1+cos \Theta)(1+cos \Theta)}{(1-cos \Theta)(1+cos \Theta)}}\)

= \(\sqrt{\frac{(1-cos \Theta)^{2}}{(1-cos^{2} \Theta)}}+\sqrt{\frac{(1+cos \Theta)^{2}}{(1-cos^{2} \Theta)}}\)

= \(\sqrt{\frac{(1-cos \Theta)^{2}}{(sin^{2} \Theta)}}+\sqrt{\frac{(1+cos \Theta)^{2}}{(sin^{2} \Theta)}}\)

= \(\frac{(1-cos \Theta)}{(sin \Theta)}+\frac{(1+cos\Theta)}{(sin \Theta)}\)

= \(\frac{(1-cos \Theta +1+cos \Theta)}{(sin \Theta)}\)

= \(\frac{(2)}{(sin \Theta)}\)

= 2cosec \Theta

Therefore, LHS = RHS

Hence proved

(iv) \(\frac{sec\Theta-1}{sec\Theta+1}\) = \((\frac{sin\Theta}{1+cos\Theta})^{2}\)

Ans:

To prove,

\(\frac{sec\Theta-1}{sec\Theta+1}\) = \((\frac{sin\Theta}{1+cos\Theta})^{2}\)

Considering left hand side (LHS),

= \(\frac{sec\Theta-1}{sec\Theta+1}\)

= \(\frac{1-cos\Theta}{1+cos\Theta}\)

Multiply and divide with (1+cos\(\Theta\))

= \(\frac{(1-cos\Theta)(1+cos\Theta)}{(1+cos\Theta)(1+cos\Theta)}\)

= \(\frac{(1-cos^{2}\Theta)}{(1+cos\Theta)^{2}}\)

= \(\frac{(sin^{2}\Theta)}{(1+cos\Theta)^{2}}\)

= \((\frac{sin\Theta}{1+cos\Theta})^{2}\)

Therefore, LHS = RHS

Hence proved

39. (sec A – tan A)² = \(\frac{1-sin A}{1 + sin A}\)

Ans:

To prove,

(sec A – tan A)² = \(\frac{1-sin A}{1 + sin A}\)

Considering left hand side (LHS),

= (sec A – tan A)²

= \([\frac{1}{cos A}-\frac{sin A}{cos A}]^{2}\)

= \(\frac{(1-sinA)^{2}}{cos^{2}A}\)

= \(\frac{(1-sinA)^{2}}{1-sin^{2}A}\)

= \(\frac{(1-sinA)^{2}}{(1+sinA)(1-sinA)}\)

= \(\frac{(1-sinA)}{(1+sinA)}\)

Therefore, LHS = RHS

Hence proved

40. \(\frac{1 – cos A}{1 + cos A}\) = (cot A – cosec A)²

Ans:

To prove,

\(\frac{1- cos A}{1 + cos A}\) = (cot A – cosecA)

Considering left hand side (LHS),

Rationalize the numerator and denominator with (1 – cos A)

= \(\frac{(1- cos A)(1- cos A)}{(1 + cos A)(1- cos A)}\)

= \(\frac{(1- cos A)^{2}}{(1- cos^{2} A)}\)

= \(\frac{(1- cos A)^{2}}{(sin^{2} A)}\)

= \((\frac{1}{sin A}-\frac{cos A}{sin A})^{2}\)

= (cosec A – cot A)²

= (cot A – cosec)²

Therefore, LHS = RHS

Hence proved

41. \(\frac{1}{sec A – 1}+\frac{1}{sec A+1} = 2 cosec A cot A\)

Ans:

To prove,

\(\frac{1}{sec A – 1}+\frac{1}{sec A+1} = 2 cosec A cot A\)

Considering left hand side (LHS),

= \(\frac{sec A+1+sec A-1}{(sec A+1)(sec A-1)}\)

= \(\frac{2sec A}{(sec^{2} A-1)}\)

= \(\frac{2sec A}{(tan^{2} A)}\)

= \(\frac{2cos^{2} A}{(cos Asin^{2} A)}\)

= \(\frac{2cos A}{(sin^{2} A)}\)

= \(\frac{2cos A}{(sin A)(sin A))}\)

= 2cosec A cot A

Therefore, LHS = RHS

Hence proved

42. \(\frac{cos A}{1-tan A}+\frac{sin A}{1-cot A}\) = sin A + cos A

Ans:

To prove,

\(\frac{cos A}{1-tan A}+\frac{sin A}{1-cot A}\) = sin A + cos A

Considering left hand side (LHS),

= \(\frac{cos A}{1-tan A}+\frac{sin A}{1-cot A}\)

= \(\frac{cos A}{1-\frac{sin A}{cos A}}+\frac{sin A}{1-\frac{cos A}{sin A}}\)

= \(\frac{cos^{2} A}{cos A-sin A}-\frac{sin^{2} A}{cos A-sin A}\)

= \(\frac{cos^{2} A-sin^{2} A}{cos A-sin A}\)

= \(\frac{(cos A + sin A)(cos A-sin A)}{cos A-sin A}\)

= cos A + sin A

Therefore, LHS = RHS

Hence proved

43. \(\frac{(cosec A)}{(cosec A-1)}+\frac{(cosec A)}{(cosec A+1)}\) = 2sec² A

Ans:

To prove,

\(\frac{(cosec A)}{(cosec A-1)}+\frac{(cosec A)}{(cosec A+1)}\) = 2sec² A

Considering left hand side (LHS),

= \(\frac{(cosec A)(cosec A+1+cosec A-1)}{(cosec^{2} A-1)})\)

= \(\frac{(2cosec^{2} A)}{cot^{2}A}\)

= \(\frac{(2sin^{2} A)}{sin^{2} A.cos^{2}A}\)

= \(\frac{2}{cos^{2}A}\)

= \(2sec^{2}A\)

Therefore, LHS = RHS

Hence proved

44.\(\frac{tan^{2}A}{1+tan^{2}A}+\frac{cot^{2}A}{1+cot^{2}A}\) = 1

Ans:

To prove,

\(\frac{tan^{2}A}{1+tan^{2}A}+\frac{cot^{2}A}{1+cot^{2}A}\) = 1

Considering left hand side (LHS),

=\(\frac{\frac{sin^{2}A}{cos^{2}A}}{\frac{cos^{2}A+sin^{2}A}{cos^{2}A}} + \frac{\frac{cos^{2}A}{sin^{2}A}}{\frac{cos^{2}A+sin^{2}A}{sin^{2}A}}\)

= \(\frac{sin^{2}A}{cos^{2}A+sin^{2}A} + \frac{cos^{2}A}{cos^{2}A+sin^{2}A}\)

= \(\frac{sin^{2}A+cos^{2}A}{cos^{2}A+sin^{2}A}\)

= 1

Therefore, LHS = RHS

Hence proved

45. \(\frac{cot A-cos A}{cot A+cos A}\) = \(\frac{cosec A-1}{cosec A + 1}\)

Ans:

To prove,

\(\frac{cot A-cos A}{cot A+cos A}\) = \(\frac{cosec A-1}{cosec A + 1}\)

Considering left hand side (LHS),

= \(\frac{\frac{cos A}{sin A}-cos A}{\frac{cos A}{sin A}+cos A}\)

= \(\frac{cosA cosecA-cos A}{cosA cosecA +cos A}\)

= \(\frac{cosA (cosecA-1)}{cosA (cosecA +1)}\)

= \(\frac{(cosecA-1)}{(cosecA +1)}\)

Therefore, LHS = RHS

Hence proved

46. \(\frac{1+cos\Theta -sin^{2}\Theta }{sin\Theta (1+cos\Theta)}\) = cot \(\Theta\)

Ans:

To prove,

\(\frac{1+cos\Theta -sin^{2}\Theta }{sin\Theta (1+cos\Theta)}\) = cot \(\Theta\)

Considering left hand side (LHS),

= \(\frac{1+cos\Theta -(1-cos^{2}\Theta) }{sin\Theta (1+cos\Theta)}\)

= \(\frac{1+cos\Theta -1+cos^{2}\Theta}{sin\Theta (1+cos\Theta)}\)

= \(\frac{cos\Theta + cos^{2}\Theta}{sin\Theta (1+cos\Theta)}\)

= \(\frac{cos\Theta(1 + cos\Theta)}{sin\Theta (1+cos\Theta)}\)

= \(\frac{(cos\Theta)}{(sin\Theta)}\)

= \(cot\Theta\)

Therefore, LHS = RHS

Hence, proved.

(i) \(\frac{1+cos\Theta +sin\Theta }{1+cos\Theta -sin\Theta }\) = \(\frac{1+sin\Theta }{cos\Theta }\)

Ans:

To prove,

\(\frac{1+cos\Theta +sin\Theta }{1+cos\Theta -sin\Theta }\) = \(\frac{1+sin\Theta }{cos\Theta }\)

Dividing the numerator and denominator with \(cos\Theta\)

Considering LHS, we get,

= \(\frac{\frac{1+cos\Theta +sin\Theta}{cos\Theta }}{\frac{1+cos\Theta -sin\Theta}{cos\Theta }}\)

= \(\frac{sec\Theta+1+tan\Theta }{sec\Theta+1-tan\Theta }\)

= \(\frac{1+sec\Theta+tan\Theta }{1+sec\Theta-tan\Theta }\)

\((sec^{2}\Theta )-(tan^{2}\Theta ) = 1\\ (sec\Theta+tan\Theta)(sec\Theta-tan\Theta ) = 1\\ (sec\Theta+tan\Theta) = \frac{1}{(sec\Theta-tan\Theta)}\)]

= \(\frac{\frac{1}{(sec\Theta-tan\Theta)}+1}{1+sec\Theta-tan\Theta}\)

= \(\frac{1+sec\Theta-tan\Theta}{1+sec\Theta-tan\Theta}\times \frac{1}{sec\Theta-tan\Theta}\)

= \(sec\Theta+tan\Theta\)

= \(\frac{1+sin\Theta }{cos\Theta }\)

Therefore, LHS = RHS

Hence proved

(ii)  \(\frac{sin\Theta-cos\Theta+1 }{sin\Theta+cos\Theta-1}\) = \(\frac{1}{sec\Theta-tan\Theta}\)

Ans:

To prove,

\(\frac{sin\Theta-cos\Theta+1 }{sin\Theta+cos\Theta-1}\) = \(\frac{1}{sec\Theta-tan\Theta}\)

Considering LHS, we get,

\(\frac{sin\Theta-cos\Theta+1 }{sin\Theta+cos\Theta-1}\)

Dividing the numerator and denominator with \(cos\Theta\), we get,

= \(\frac{tan\Theta+sec\Theta-1 }{tan\Theta-sec\Theta+1}\)

= \(\frac{ \frac{1}{(sec\Theta-tan\Theta)}-1}{tan\Theta-sec\Theta+1}\)

= \(\frac{tan\Theta-sec\Theta+1}{tan\Theta-sec\Theta+1}\times \frac{1}{(sec\Theta-tan\Theta)}\)

= \(\frac{1}{(sec\Theta-tan\Theta)}\)

Therefore, LHS = RHS

Hence proved

(iii) \(\frac{cos\Theta-sin\Theta+1 }{cos\Theta+sin\Theta-1}\) = \(cosec\Theta + cot\Theta\)

Ans:

To prove,

\(\frac{cos\Theta-sin\Theta+1 }{cos\Theta+sin\Theta-1}\) = \(cosec\Theta + cot\Theta\)

Considering LHS, we get,

Dividing the numerator and denominator with \(sin\Theta\), we get,

= \(\frac{\frac{cos\Theta-sin\Theta+1}{sin\Theta } }{\frac{cos\Theta+sin\Theta-1}{sin\Theta}}\)

= \(\frac{cot\Theta+cosec\Theta-1}{cot\Theta-cosec\Theta+1}\)

= \(\frac{\frac{1}{(cosec\Theta-cot\Theta)}-1}{cot\Theta-cosec\Theta+1}\)

= \(\frac{cot\Theta-cosec\Theta+1}{cot\Theta-cosec\Theta+1}\times \frac{1}{(cosec\Theta-cot\Theta)}\)

= \(\frac{1}{(cosec\Theta-cot\Theta)}\)

= \(cosec\Theta + cot\Theta\)

Therefore, LHS = RHS

Hence proved

(iv) \((sin\Theta + cos\Theta )(tan\Theta + cot\Theta)\) = \(sec\Theta+cosec\Theta\)

Ans:

To prove,

\((sin\Theta + cos\Theta )(tan\Theta + cot\Theta)\) = \(cosec\Theta+cosec\Theta\)

Considering LHS, we get,

= \((sin\Theta + cos\Theta)(\frac{sin\Theta}{cos\Theta }+\frac{cos\Theta}{sin\Theta})\)

= \((\frac{sin^{2}\Theta}{cos\Theta}+ cos\Theta+sin\Theta+\frac{cos^{2}\Theta}{sin\Theta}\)

= \(sin\Theta(tan\Theta +1) +cos\Theta( \frac{1}{tan\Theta}+1)\)

= \(sin\Theta(tan\Theta +1) +\frac{cos\Theta}{tan\Theta}(tan\Theta +1)\)

= \((sin\Theta+\frac{cos\Theta}{tan\Theta})(tan\Theta +1)\)

= \((\frac{sin^{2}\Theta+cos^{2}\Theta}{sin\Theta})(tan\Theta +1)\)

= \((\frac{1}{sin\Theta})(tan\Theta +1)\)

= \((\frac{sin\, \theta +cos\Theta }{cos\Theta \; sin\Theta })\)

= \(sec\Theta+cosec\Theta\)

Therefore, LHS = RHS

Hence proved

50. \(\frac{tanA}{1+secA}-\frac{tanA}{1-secA}\)= 2 cosec A

Ans:

To prove,

Considering LHS, we get,

= \(\frac{\frac{sin A}{cos A}}{\frac{cosA+1}{cosA}}-\frac{\frac{sin A}{cos A}}{\frac{cosA-1}{cosA}}\)

= \(\frac{sinA}{cosA+1}-\frac{sinA}{cosA-1}\)

= \(sinA(\frac{1}{cosA+1}-\frac{1}{cosA-1})\)

= \(sinA(\frac{cosA-1-cosA-1}{cos^{2}A-1})\)

= \(sinA(\frac{cosA-1-cosA-1}{cos^{2}A-1})\)

= \(sinA(\frac{-2}{-sin^{2}A})\)

= \(\frac{2}{sinA})\)

= 2 cosec A

∴ LHS = RHS

Hence proved

Q51: \(1+\frac{cot^{2}\Theta}{1+cosec\Theta}=cosec\Theta\)

Ans:

\(1+\frac{cosec^{2}\Theta-1}{1+cosec\Theta}\;\;\;\;[because\; cot^{2}\Theta=cosec^{2}\Theta-1]\)

\(1+\frac{(cosec\Theta-1)(cosec\Theta+1)}{1+cosec\Theta}\)

\(=1+cosec\Theta-1\;\;\;\;\;[because\; (a+b)(a-b)=a^{2}-b^{2})]\)

\(=cosec\Theta\)

∴ LHS = RHS

Hence, proved.

Q52:\(\frac{cos\Theta}{cosec\Theta+1}+\frac{cos\Theta}{cosec\Theta-1}=2tan\Theta\)

Ans:

\(\frac{cos\Theta}{\frac{1}{sin\Theta}+1}+\frac{cos\Theta}{\frac{1}{sin\Theta}-1}\)

\(\frac{cos\Theta}{\frac{1+sin\Theta}{sin\Theta}}+\frac{cos\Theta}{\frac{1-sin\Theta}{sin\Theta}}\)

\(\frac{(cos\Theta)(sin\Theta)}{1+sin\Theta}+\frac{(cos\Theta)(sin\Theta)}{1-sin\Theta}\)

\(\frac{(1-sin\Theta)(sin\Theta cos\Theta)+(sin\Theta cos\Theta)}{(1+sin\Theta)(1-sin\Theta)}\)

\(\frac{sin\Theta cos\Theta-sin\Theta cos\Theta+sin\Theta cos\Theta+sin^{2}\Theta cos^{2}\Theta}{1-sin^{2}\Theta}\)

\(=\frac{sin\Theta cos\Theta}{cos^{2}\Theta}\)

\(=\frac{2sin\Theta}{cos\Theta}\)

\(=2tan\Theta\)

Therefore, LHS = RHS

Hence, proved

Q53) \((1+tan^{2}A)+(1+\frac{1}{tan^{2}A})=\frac{1}{sin^{2}A-sin^{4}A}\)

Ans:

LHS = \((1+\frac{sin^{2}A}{cos^{2}A})+(1+\frac{cos^{2}A}{sin^{2}A})\)

=>\(\frac{cos^{2}A+sin^{2}A}{cos^{2}A}+\frac{sin^{2}A+cos^{2}A}{sin^{2}A}\)

=>\(\frac{1}{cos^{2}A}+\frac{1}{sin^{2}A}\;\;\;\;\;\;[because\; sin^{2}A+cos^{2}A=1]\)

=>\(\frac{sin^{2}A+cos^{2}A}{sin^{2}Acos^{2}A}=\frac{1}{sin^{2}A(1-sin^{2}A)}\;\;\;\;\;\;[because\; cos^{2}A=1-sin^{2}A]\)

=>\(\frac{1}{sin^{2}A-sin^{4}A}\)

∴ LHS = RHS.

Hence Proved.

Q54) sin²Acos²B – cos²Asin²B = sin²A – sin²B

Ans:

LHS = sin²Acos²B – cos²Asin²B

= \(sin^{2}A(1-sin^{2}B)-(1-sin^{2}A)(sin^{2}A)\;\;\;\;\;\;[because\, cos^{2}A=1-sin^{2}A]\)

= \(sin^{2}A-sin^{2}Asin^{2}B-sin^{2}B+sin^{2}Asin^{2}B\)

= \(sin^{2}A-sin^{2}B\)

= RHS

Hence Proved.

Q55: (i) \(\frac{cotA+tanB}{cotB+tanA}=cotAtanB\)

Ans:

LHS = \(\frac{cotA+tanB}{cotB+tanA}\)

= \(\frac{\frac{cosA}{sinA}+\frac{sinB}{cosB}}{\frac{cosB}{sinB}+\frac{sinA}{cosA}}\)

= \(\frac{\frac{cosAcosB+sinAsinB}{sinAcosB}}{\frac{cosAcosB+sinAsinB}{cosAsinB}}\)

= \(\frac{cosAcosB+sinAsinB}{sinAcosB}\times \frac{cosAsinB}{cosAcosB+sinAsinB}\)

= \(\frac{cosAsinB}{sinAcosB}\)

= cotAtanB

= RHS

Hence Proved.

 (ii) \(\frac{tanA+tanB}{cotA+cotB}=tanAtanB\)

Ans:

LHS = \(\frac{tanA+tanB}{cotA+cotB}\)

= \(\frac{\frac{sinA}{cosA}+\frac{sinB}{cosB}}{\frac{cosA}{sinA}+\frac{cosB}{sinB}}\)

= \(\frac{\frac{sinAcosB+cosAsinB}{cosAcosB}}{\frac{cosAsinB+cosBsinA}{sinAsinB}}\)

= \(\frac{sinAcosB+cosAsinB}{cosAcosB}\times \frac{sinAsinB}{cosAsinB+cosBsinA}\)

= \(\frac{sinAsinB}{cosAcosB}\)

= tanAtanB

= RHS

Hence Proved.

Q56) \(cot^{2}Acosec^{2}B-cot^{2}Bcosec^{2}A=cot^{2}A-cot^{2}B\)

Ans:

LHS = \(cot^{2}Acosec^{2}B-cot^{2}Bcosec^{2}A\)

= \(cot^{2}A(1+cot^{2}B)-cot^{2}B(1+cot^{2}A)\;\;\;\;\;\;[because\, cosec^{2}\theta=1+cot^{2}\theta]\)

= \(cot^{2}A+cot^{2}Acot^{2}B-cot^{2}B-cot^{2}Bcot^{2}A\)

= cot²A – cot²B

= RHS

Hence Proved.

Q57) \(tan^{2}Asec^{2}B-sec^{2}Atan^{2}B=tan^{2}A-tan^{2}B\)

Ans:

LHS = \(tan^{2}Asec^{2}B-sec^{2}Atan^{2}B\)

= \(tan^{2}A(1+tan^{2}B)-sec^{2}A(tan^{2}A)\)

= \(tan^{2}A+tan^{2}Atan^{2}B-tan^{2}B(1+tan^{2}A)\;\;\;\;\;[because\, sec^{2}A=1+tan^{2}A]\)

= \(tan^{2}A+tan^{2}Atan^{2}B-tan^{2}B-tan^{2}Atan^{2}B\)

= \(tan^{2}A-tan^{2}B\)

= RHS

Hence Proved.

Q58) If x = \(asec\theta+btan\theta\) and y = \(a\;tan\theta+b\;sec\theta\), prove that x² – y² = a² – b².

Ans:

LHS = x² – y²

= \((asec\theta+btan\theta)^{2}-(atan\theta+bsec\theta)^{2}\)

= \(a^{2}sec^{2}\theta+b^{2}tan^{2}\theta+2absec\theta tan\theta-a^{2}tan^{2}\theta-b^{2}sec^{2}\theta-2absec\theta tan\theta\)

= \(a^{2}sec^{2}\theta+b^{2}tan^{2}\theta-a^{2}tan^{2}\theta-b^{2}sec^{2}\theta\)

= \(a^{2}sec^{2}\theta-b^{2}sec^{2}\theta+b^{2}tan^{2}\theta-a^{2}tan^{2}\theta\)

= \(sec^{2}\theta(a^{2}-b^{2})+tan^{2}\theta(b^{2}-a^{2})\)

= \(sec^{2}\theta(a^{2}-b^{2})-tan^{2}\theta(a^{2}-b^{2})\)

= \((sec^{2}\theta-tan^{2}\theta)(a^{2}-b^{2})\)

= a² – b²

= RHS

Hence Proved.

Q59) If \(3sin\theta+5cos\theta=5\), prove that \(5sin\theta-3cos\theta=\pm 3\).

Ans:

Given \(3sin\theta+5cos\theta=5\)

\(3sin\theta=5-5cos\theta\)

\(3sin\theta=5(1-cos\theta)\)

\(3sin\theta=\frac{5(1-cos\theta)(1-cos\theta)}{1+cos\theta}\)

\(3sin\theta=\frac{5(1-cos^{2}\theta)}{1+cos\theta}\)

\(3sin\theta=\frac{5sin^{2}\theta}{1+cos\theta}\)

\(3+3cos\theta=5sin\theta\)

\(3=5sin\theta-3cos\theta\)

= RHS

Hence Proved.

Q60) If \(cosec\theta+cot\theta\)=m and \(cosec\theta-cot\theta\)=n, prove that mn = 1.

Ans:

LHS = mn

= \((cosec\theta+cot\theta)(cosec\theta-cot\theta)\)

= \(cosec^{2}\theta-cot^{2}\theta\)

= 1

= RHS

Hence Proved.

Q 62 . If \(T_{n}=sin^{n}\theta +cos_{n}\theta\) , prove that \(\frac{T_{3}-T_{5}}{T_{1}}= \frac{T_{5}-T_{7}}{T_{3}}\) .

Ans:

LHS = \(\frac{\left (sin^{3}\theta +cos^{3}\theta \right ) – \left (sin^{5}\theta +cos^{5}\theta \right ) }{sin\theta +cos\theta }\)

= \(\frac{sin^{3}\theta \left ( 1- sin^{2}\theta \right )+cos^{3}\theta \left ( 1- cos^{2}\theta \right ) }{sin\theta +cos\theta }\)

= \(\frac{sin^{3}\theta \times cos^{2}\theta +cos^{3}\theta \times sin^{2}\theta }{sin\theta +cos\theta }\)

= \(\frac{sin^{2}\theta cos^{2}\theta \left ( sin\theta +cos\theta \right ) }{sin\theta +cos\theta }\)

= \(sin^{2}\theta cos^{2}\theta\)

RHS = \(\frac{\left (sin^{5}\theta +cos^{5}\theta \right ) – \left (sin^{7}\theta +cos^{7}\theta \right ) }{sin^{3}\theta +cos^{3]\theta }\)

= \(\frac{sin^{5}\theta \left ( 1- sin^{2}\theta \right )+cos^{5}\theta \left ( 1- cos^{2}\theta \right ) }{sin^{3}\theta +cos^{3}\theta }\)

= \(\frac{sin^{5}\theta \times cos^{2}\theta +cos^{5}\theta \times sin^{2}\theta }{sin^{3}\theta +cos^{3}\theta }\)

= \(\frac{sin^{2}\theta cos^{2}\theta \left ( sin^{3}\theta +cos^{3}\theta \right ) }{sin\theta +cos\theta }\)

= \(sin^{2}\theta cos^{2}\theta\)

∴ LHS = RHS   Hence proved .

Q 63 . \(\left ( tan\theta +\frac{1}{cos\theta } \right )^{2}+\left ( tan\theta -\frac{1}{cos\theta } \right )^{2}\) = \(2\left ( \frac{1+sin^{2}\theta }{1-sin^{2}\theta } \right )\)

Ans:

\(\left ( tan \theta +sec\theta \right )^{2}+\left ( tan \theta -sec\theta \right )^{2}\)

= \(tan ^{2}\theta +sec^{2}\theta +2 tan\theta sec\theta +tan^{2} \theta +sec^{2}\theta -2tan\theta sec\theta\)

= \(2tan ^{2}\theta +2sec^{2}\theta\)

= \(2\left [tan ^{2}\theta +sec^{2}\theta \right ]\)

= \(2\left [\frac{sin^{2}\theta }{cos^{2}\theta} +\frac{1}{cos^{2}\theta} \right ]\)

= \(2\left ( \frac{1+sin^{2}\theta }{cos^{2}\theta } \right )\)

= \(2\left ( \frac{1+sin^{2}\theta }{1-sin^{2}\theta } \right )\)

= RHS

∴ LHS = RHS   Hence proved .

Q 64 . \(\left (\frac{1}{sec^{2}\theta -cos^{2}\theta}+\frac{1}{cosec^{2}\theta -sin^{2}} \right )sin^{2}\theta cos^{2}\theta\) = \(\frac{1-sin^{2}\theta cos^{2}\theta }{2+sin^{2}\theta cos^{2}\theta }\)

Ans:

\(\left [ \frac{1}{\frac{1}{cos^{2}\theta }-cos^{2}\theta } +\frac{1}{\frac{1}{sin^{2}\theta }-sin^{2}\theta}\right ]sin^{2}\theta cos^{2}\)

= \(\left [ \frac{1}{\frac{1-cos^{4}\theta}{cos^{2}\theta } } +\frac{1}{\frac{1-sin^{4}\theta }{sin^{2}\theta}}\right ]sin^{2}\theta cos^{2}\theta \)

= \(\left [ \frac{cos^{2}\theta }{1-cos^{4}\theta} +\frac{sin^{2}\theta}{1-sin^{4}\theta}\right ]sin^{2}\theta cos^{2}\theta\)

= \(\left [ \frac{cos^{2}\theta }{cos^{2}\theta+sin^{2}\theta-cos^{4}\theta} +\frac{sin^{2}\theta}{cos^{2}\theta+sin^{2}\theta-sin^{4}\theta}\right ]sin^{2}\theta cos^{2}\theta\)

= \(\left [ \frac{cos^{2}\theta }{cos^{2}\theta\left ( 1-cos^{2}\theta \right )+sin^{2}\theta} +\frac{sin^{2}\theta}{cos^{2}\theta+sin^{2}\theta\left ( 1- sin^{2}\theta\right )}\right ]sin^{2}\theta cos^{2}\theta\)

= \(\left [ \frac{cos^{2}\theta }{cos^{2}\theta sin^{2}\theta+sin^{2}\theta} +\frac{sin^{2}\theta}{cos^{2}\theta+sin^{2}\theta cos^{2}\theta}\right ]sin^{2}\theta cos^{2}\theta\)

= \(\left [ \frac{cos^{2}\theta }{sin^{2}\theta \left ( cos^{2}\theta +1 \right )} +\frac{sin^{2}\theta}{cos^{2}\theta\left ( sin^{2}\theta+1 \right )}\right ]sin^{2}\theta cos^{2}\theta\)

= \(\frac{cos^{4}\theta \left ( sin^{2}\theta+1 \right )+sin^{4}\theta\left ( cos^{2}\theta +1 \right )}{sin^{2}\theta cos^{2}\theta \left ( cos^{2}\theta +1 \right )\left (sin^{2}\theta+1 \right ) } sin^{2}\theta cos^{2}\theta\)

= \(\frac{cos^{4}\theta \left ( sin^{2}\theta+1 \right )+sin^{4}\theta\left ( cos^{2}\theta +1 \right )}{ \left ( cos^{2}\theta +1 \right )\left (sin^{2}\theta+1 \right ) }\)

= \(\frac{cos^{4}\theta +cos^{4}\theta sin^{2}\theta +sin^{4}\theta +sin^{4}\theta cos^{2}\theta }{1 +sin^{2}\theta +cos^{2}\theta +cos^{2}\theta sin^{2}\theta}\)

= \(\frac{1-2sin^{2}\theta cos^{2}\theta+ sin^{2}\theta cos^{2}\theta\left ( cos^{2}\theta +sin^{2}\theta \right ) }{1 +1+cos^{2}\theta sin^{2}\theta}\)

= \(\frac{1-sin^{2}\theta cos^{2}\theta }{2+sin^{2}\theta cos^{2}\theta }\)

∴ LHS = RHS   Hence proved .

Q 65 . (i) . \(\left [ \frac{1+sin\theta -cos\theta }{1+sin\theta +cos\theta} \right ]^{2}\) = \(\frac{1-cos\theta }{1+cos\theta}\)

Ans:

= \(\left ( \frac{1+sin\theta -cos\theta }{1+sin\theta +cos\theta}\times \frac{1+sin\theta -cos\theta}{1+sin\theta -cos\theta} \right )^{2}\)

= \(\left [ \frac{\left (1+sin\theta -cos\theta \right )^{2}}{\left ( 1+sin\theta \right )^{2}-cos^{2}\theta } \right ]^{2}\)

= \(\left [ \frac{\left ( 1 \right )^{2}+sin^{2}\theta +cos^{2}\theta +2\times 1\times sin\theta +2\times sin\theta \left ( -cos\theta \right )-2cos\theta }{1-cos^{2}\theta+sin^{2}\theta +2 sin\theta} \right ]^{2}\)

= \(\left [ \frac{ 1 +1 +2 sin\theta -2 sin\theta cos\theta -2cos\theta }{sin^{2}\theta+sin^{2}\theta +2 sin\theta} \right ]^{2}\)

= \(\left [ \frac{ 2 +2 sin\theta -2 sin\theta cos\theta -2cos\theta }{2sin^{2}\theta +2 sin\theta} \right ]^{2}\)

= \(\left [ \frac{ 2 \left (1+ sin\theta \right )-2cos\theta \left (sin\theta +1 \right ) }{2sin\theta \left (sin\theta+1 \right )} \right ]^{2}\)

= \(\left [ \frac{ \left (1+ sin\theta \right ) \left (2-2cos\theta \right ) }{2sin\theta \left (sin\theta+1 \right )} \right ]^{2}\)

= \(\left [ \frac{ \left (2-2cos\theta \right ) }{2sin\theta } \right ]^{2}\)

= \(\left [ \frac{ \left (1-cos\theta \right ) }{sin\theta } \right ]^{2}\)

= \(\frac{ \left (1-cos\theta \right )^{2} }{1-cos^{2}\theta }\)

= \(\frac{ \left (1-cos\theta \right )\times \left ( 1-cos\theta \right ) }{\left (1-cos\theta \right ) \left ( 1+cos\theta \right )}\)

= \(\frac{1-cos\theta }{1+cos\theta}\)

∴ LHS = RHS   Hence proved .

Q 65 (ii) . \(\frac{1+sec\theta -tan\theta }{1+sec\theta +tan\theta}\) = \(\frac{1-sin\theta }{cos\theta }\)

Ans:

= LHS =  \(\frac{1+sec\theta -tan\theta }{1+sec\theta +tan\theta}\)

= \(\frac{\left ( sec^{2 } \theta -tan^{2}\theta\right )+\left ( sec\theta -tan\theta \right )}{1+sec\theta +tan\theta }\)                               [since , \(sec^{2}\theta -tan^{2}\theta = 1\)]

= \(\frac{\left ( sec \theta -tan\theta\right )\left ( sec\theta +tan\theta \right )+\left ( sec\theta -tan\theta \right )}{1+sec\theta +tan\theta }\)

= \(\frac{\left ( sec \theta -tan\theta\right )\left (1+sec\theta +tan\theta \right )}{1+sec\theta +tan\theta }\)

= \(\left ( sec \theta -tan\theta\right )\)

= \(\frac{1}{cos\theta} -\frac{sin\theta}{cos\theta}\)

= \(\frac{1-sin\theta }{cos\theta }\)

∴ LHS = RHS   Hence proved .

Q 66 . ( sec A + tan A – 1 )( sec A – tan A + 1) = 2 tan A

Ans:

= \(\left ( sec A+tan A-\left \{ sec^{2}A-tan^{2}A \right \} \right )\left [sec A-tan A +\left ( sec^{2}A-tan^{2}A \right ) \right ]\)

= \(\left ( sec A+tan A-\left ( sec A+tan A \right )\left (sec A-tan A \right ) \right )\left [sec A-tan A +\left ( sec A-tan A \right )\left (sec A+tan A \right ) \right ]\)

= \(\left ( secA+tanA \right )\left (1-\left (secA-tanA \right ) \right )\left ( secA-tanA \right )\left (1+\left (secA+tanA \right ) \right )\)

= \(\left ( sec^{2}A-tan^{2}A \right )\left (1-secA+tanA \right )\left (1+secA+tanA \right )\)

= \(\left (1-\frac{1}{cos A}+\frac{sin A}{cos A} \right )\left (1+\frac{1}{cos A}+\frac{sin A}{cos A} \right )\)

= \(\left (\frac{cos A – 1+sin A}{cos A} \right )\left (\frac{cos A +1+sin A}{cos A} \right )\)

= \(\left (\frac{\left (cos A+sin A \right )^{2}-1}{cos^{2} A} \right )\)

= \(\left (\frac{cos^{2} A+sin ^{2}A +2 sin A cos B -1}{cos^{2} A} \right )\)

= \(\left (\frac{1 +2 sin A cos B -1}{cos^{2} A} \right )\)

= \(\left (\frac{2 sin A cos B }{cos^{2} A} \right )\)

= 2 tan A

∴ LHS = RHS   Hence proved .

Q 67 . ( 1 + cot A – cosec A )( 1 + tan A + sec A ) = 2

Ans:

LHS = ( 1 + cot A – cosec A )( 1 + tan A + sec A )

= \(\left ( 1+\frac{cos A}{sin A}-\frac{1}{sin A} \right )\left (1+\frac{sin A}{cos A}+\frac{1}{cos A} \right )\)

= \(\left ( \frac{sin A+cos A-1}{sin A} \right )\left (\frac{cos A+sin A+1}{cos A} \right )\)

= \(\left ( \frac{\left (sin A-cos A \right )^{2}-1}{sin Acos A} \right )\)

= \(\frac{sin^{2}A+2sinAcosA+cos^{2}A-1}{sinAcosA}\)

= \(\left ( \frac{1+2sin A cos A-1}{sin A cos A} \right )\)

= 2

∴ LHS = RHS   Hence proved .

Q 68 . \(\left ( cosec\theta -sec\theta \right )\left ( cot\theta -tan\theta \right )\) = \(\left ( cosec\theta +sec\theta \right )\left ( sec\theta cosec\theta – 2 \right )\)

Ans:

LHS =  \(\left ( cosec\theta -sec\theta \right )\left ( cot\theta -tan\theta \right )\)

\(\left [ \frac{1}{sin\theta }- \frac{1}{cos\theta }\right ]\left [ \frac{cos\theta }{sin\theta }-\frac{sin\theta }{cos\theta } \right ]\)

\(\left [ \frac{cos\theta -sin\theta }{sin\theta cos\theta }\right ]\left [ \frac{cos^{2}\theta -sin^{2}\theta}{sin\theta cos\theta } \right ]\)

\(\left [ \frac{\left (cos\theta -sin\theta \right )^{2}\left ( cos\theta +sin\theta \right )}{sin^{2}\theta cos^{2}\theta }\right ]\)

RHS  = \(\left ( cosec\theta +sec\theta \right )\left ( sec\theta cosec\theta – 2 \right )\)

\(\left [ \frac{1}{sin\theta }+ \frac{1}{cos\theta }\right ]\left [ \frac{1 }{cos\theta }-\frac{1 }{sin\theta }-2 \right ]\)

= \(\left [ \frac{sin\theta +cos\theta }{sin\theta cos\theta} \right ]\left [ \frac{1-2 sin\theta cos\theta}{sin\theta cos\theta} \right ]\)

= \(\left [ \frac{sin\theta +cos\theta }{sin\theta cos\theta} \right ]\left [ \frac{cos^{2}\theta +sin^{2}\theta -2 sin\theta cos\theta}{sin\theta cos\theta} \right ]\)

= \(\left [ \frac{\left (cos\theta -sin\theta \right )^{2}\left ( cos\theta +sin\theta \right )}{sin^{2}\theta cos^{2}\theta }\right ]\)                                                 \(\left [ because\, cos^{2}\theta +sin^{2}\theta = 1 \right ]\)

∴ LHS = RHS   Hence proved .

Q 70 . \(\frac{cos A cosec A-sin A sec A}{cos A +sin A}\) = cosec A – sec A

Ans:

LHS = \(\frac{cos A cosec A-sin A sec A}{cos A +sin A}\)

= \(\frac{cos A \times \frac{1}{sin A}-sin A \times \frac{1}{cos A}}{cos A +sin A}\)

= \(\frac{\frac{cos A}{sin A}- \frac{sin A}{cos A}}{cos A +sin A}\)

= \(\frac{\frac{cos^{2} A- sin^{2}A}{cos A sin A}}{cos A +sin A}\)

= \(\frac{cos^{2} A- sin^{2}A}{cos A sin A}\times \frac{1}{cos A +sin A}\)

= \(\frac{\left (cos A- sinA \right )\left ( cos A+ sinA \right )}{cos A sin A\times\left ( cos A +sin A \right )}\)

= \(\frac{\left (cos A- sinA \right )}{cos A sin A}\)

= \(\frac{cos A }{cos A sin A} -\frac{sin A}{cos A sin A}\)

= \(\frac{1 }{ sin A} -\frac{1}{cos A }\)

= \(cosec A -sec A\)

= RHS

∴ LHS = RHS   Hence proved .

Q 71 . \(\frac{sin A}{sec A +tan A -1}+\frac{cos A}{cosec A + cot A-1}\) = 1

Ans:

LHS : \(\frac{sin A}{sec A +tan A -1}+\frac{cos A}{cosec A + cot A-1}\)

= \(\frac{sin A}{\frac{1}{cos A} +\frac{sin A}{cos A} -1}+\frac{cos A}{\frac{1}{sin A} + \frac{cos A}{sin  A}-1}\)

= \(\frac{sin A}{\frac{1+sin A – cos A}{cos A} }+\frac{cos A}{\frac{1+cos A – sin A}{sin A}}\)

= \(\frac{sin A cos A}{1+sin A – cos A}+\frac{cos A sin A}{1+cos A – sin A}\)

= \(\left ( sin A cos A \right )\left [\frac{1}{1+sin A – cos A}+\frac{1}{1+cos A – sin A} \right ]\)

= \(\left ( sin A cos A \right )\left [\frac{2}{cos A- sin A +sin A+sin A cos A- sin^{2}A-cos A-cos^{2}A+cos A sin A} \right ]\)

= \(\left ( sin A cos A \right )\left [\frac{2}{1- sin^{2}A-cos^{2}A+2sin A cos A} \right ]\)

= \(\left ( sin A cos A \right )\left [\frac{2}{1- \left (s in^{2}A-cos^{2}A \right )+2sin A cos A} \right ]\)

= \(\left ( sin A cos A \right )\left [\frac{2}{1- 1+2sin A cos A} \right ]\)

= \(\left ( sin A cos A \right )\times \frac{2}{2sin A cos A}\)

= 1

= RHS

∴ LHS = RHS   Hence proved .

Q 72 . \(\frac{tan A}{\left (1 +tan^{2}A \right )^{2}}+\frac{cot A}{\left (1 +cot^{2}A \right )^{2}}\) = sin A cos A

Ans:

\(\frac{tan A}{\left (sec^{2}A \right )^{2}}+\frac{cot A}{\left (cosec^{2}A \right )^{2}}\)                            [1 +tan² A = sec² A , 1 +cot² A = cosec² A ]

= \(\frac{\frac{sin A}{cos A}}{sec^{4}A }+\frac{\frac{cos A}{sin A}}{cosec^{4}A }\)

= \(\frac{\frac{sin A}{cos A}}{\frac{1}{cos^{4}A} }+\frac{\frac{cos A}{sin A}}{\frac{1}{sin^{4}A} }\)

= \(\frac{sin A}{cosA}\times \frac{cos^{4}A}{1}+\frac{cos A }{sin A}\times \frac{sin ^{4}A}{1}\)

= \(sin A\times cos^{3}A+cos A \times sin ^{3}A\)

= \(sin A cosA\left ( cos^{2}A+sin ^{2}A \right )\)

= \(sin A cosA\)

∴ LHS = RHS   Hence proved .

Q73. \(sec^{4}A(1\;-\;sin^{4}A)\;-\;2tan^{2}A = 1\)

Ans:

Given, L.H.S = \(sec^{4}A(1\;-\;sin^{4}A)\;-\;2tan^{2}A

= \(sec^{4}A\;-\;sec^{4}A\;\times \;sin^{4}A\;-\;2tan^{4}A\)

= \(sec^{4}A\;-\;sec^{4}A\;\times \;sin^{4}A\;-\;2tan^{4}A\]”>

= \(sec^{4}A\;-\;(\frac{1}{cos^{4}A}\;\times \;sin^{4}A)\;-\;2tan^{4}A\)

= \(sec^{4}A\;-\;tan^{4}A\;-\;2tan^{4}A\)

= \((sec^{2}A)^{2}\;-\;tan^{4}A\;-\;2tan^{4}A\)

= \((1\;+\;tan^{2}A)^{2}\;-\;tan^{4}A\;-\;2tan^{4}A\)

= \(1+tan^{4}A+2tan^{2}A-\;tan^{4}A\;-\;2tan^{4}A\)

= 1

Hence, L.H.S = R.H.S

Q74. \(\frac{cot^{2}A(secA\;-\;1)}{1\;+\;sinA}\) = \(sec^{2}A[\frac{1\;-\;sinA}{1\;+\;sinA}]\)

Ans:

Given, L.H.S = \(\frac{cot^{2}A(secA\;-\;1)}{1\;+\;sinA}\)

Here, \(sin^{2}A\;+\;cos^{2}A\) = 1

= \(\frac{\frac{cos^{2}A}{sin^{2}A}(\frac{1}{cosA}-1)}{1+sinA}\)

= \(\frac{\frac{cos^{2}A}{sin^{2}A}(\frac{1-cosA}{cosA})}{1+sinA}\)

= \(\frac{\frac{cosA\times cosA}{(1-cos^{2}A)}(\frac{1-cosA}{cosA})}{1+sinA}\)

= \(\frac{(cosA)}{(1+cosA)}\frac{1}{1+sinA}\)

Solving,

RHS =>  \(sec^{2}a[\frac{1-sinA}{1+secA}]\)

= \(\frac{1}{cos^{2}A}[\frac{1-sinA}{1+secA}]\)

= \(\frac{1}{cos^{2}A}[\frac{1-sinA}{1+secA}]\)

= \(\frac{1}{cos^{2}A}[\frac{1-sinA}{cosA+1}]cosA\)

= \(\frac{(1-sinA)}{(cosA+1)(cosA)}\)

Multiplying Nr. And Dr. with (1+SinA)

= \(\frac{(1-sinA)}{(cosA+1)(cosA)}\times\frac{1+sinA}{1+sinA}\)

= \(\frac{(1^{2}-sin^{2}A)}{(cosA+1)(cosA)(1+sinA)}\)

= \(\frac{cos^{2}A}{(cosA+1)(cosA)(1+sinA)}\)

= \(\frac{cosA}{(cosA+1)(1+sinA)}\)

Hence, LHS= RHS

Q75. \((1\;+\;cotA\;+tanA)(sinA\;-\;cosA)\) = \(\frac{secA}{cosec^{2}A}\) – \(\frac{cosecA}{sec^{2}A}\) = sinAtanA – cotAcosA

Ans:

Given, L.H.S = \((1\;+\;cotA\;+tanA)(sinA\;-\;cosA)\)

=> sinA – cosA + cotAsinA – cotAcosA + sinAtanA – tanAcosA

=> sinA – cosA + \(\frac{cosA}{sinA}\times sinA\) – cotAcosA + sinAtanA – \(\frac{sinA}{cosA}\times cosA\)

=> sinA – cosA + cosA – cotAcosA + sinAtanA – sinA

=> sinAtanA – cosAcotA

=> \(\frac{secA}{cosec^{2}A}\) – \(\frac{cosecA}{sec^{2}A}\)

Here, secA = \(\frac{1}{cosA}\) and cosecA = \(\frac{1}{sinA}\)

=> \(\frac{sin^{2}A}{cosA}\) – \(\frac{cos^{2}A}{sinA}\)

=> \(\frac{sin^{2}A\;-\;cos^{2}A}{cosAsinA}\)

=> \((sinA\times \frac{sinA}{cosA})\) – \((cosA\times \frac{cosA}{cotA})\)

=> sinAtanA – cosAcotA

Hence, L.H.S = R.H.S

Q76. If \(\frac{x}{a}cos\theta \;+\;\frac{y}{b}sin\theta\) = 1 and \(\frac{x}{a}cos\theta \;-\;\frac{y}{b}sin\theta\) = 1, prove that \(\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}\) = 2

Ans:

Given,

=> (\(\frac{x}{a}cos\theta \;+\;\frac{y}{b}sin\theta\))² + (\(\frac{x}{a}cos\theta \;-\;\frac{y}{b}sin\theta\))²         = 1² + 1²

=> \(\frac{x^{2}}{a^{2}}cos^{2}\theta \;+\;\frac{y^{2}}{b^{2}}sin^{2}\theta \;+\;\frac{2xy}{ab}cos\theta sin\theta \;+\;\frac{x^{2}}{a^{2}}sin^{2}\theta \;+\;\frac{y^{2}}{b^{2}}\;-\;\frac{2xy}{ab}sin\theta cos\theta\)        = 1 + 1

=> \(\frac{x^{2}}{a^{2}}cos^{2}\theta \;+\;\frac{y^{2}}{b^{2}}sin^{2}\theta \;+\;\frac{x^{2}}{a^{2}}sin^{2}\theta \;+\;\frac{y^{2}}{b^{2}}sin^{2}\theta\)  = 2

=> \(cos^{2}\theta [\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}]\) + \(sin^{2}\theta [\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}]\)  = 2

=> \((cos^{2}\theta\;+\;sin^{2}\theta ) [\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}]\) = 2

Here cos²A +sin²A = 1

=> (1) [\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}]\) = 2

=> [\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}]\) = 2

Q77. If \(cosec\theta \;-\;sin\theta \;=\;a^{3}\), \(sec\theta \;-\;cos\theta \;=\;b^{3}\), prove that \(a^{2}b^{2}(a^{2}\;+\;b^{2})\) = 1

Ans:

Given, \(cosec\theta \;-\;sin\theta \;=\;a^{3}\)

Here, \(cosec\theta \;=\;\frac{1}{sin\theta }\)

=> \(\frac{1}{sin\theta }\) – \(sin\theta\) = \(a^{3}\)

=> \(\frac{1\;-\;sin^{2}\theta }{sin\theta }\) = \(a^{3}\)

Here cos²A +sin²A = 1

=> \(\frac{cos^{2}\theta }{sin\theta }\) = \(a^{3}\)

=> \(\frac{cos^{\frac{2}{3}}\theta }{sin^{\frac{1}{3}}\theta }\) = a

Squaring on both sides

=> a² = \(\frac{cos^{\frac{4}{3}}\theta }{sin^{\frac{2}{3}}\theta }\)

\(sec\theta \;-\;cos\theta \;=\;b^{3}\)

=> \(\frac{1}{cos\theta }\) – \(cos\theta\) = \(b^{3}\)

=> \(\frac{1\;-\;cos^{2}\theta }{cos\theta }\) = \(b^{3}\)

=> \(\frac{sin^{2}\theta }{cos\theta }\) = \(b^{3}\)

=> \(\frac{sin^{\frac{2}{3}}\theta }{cos^{\frac{1}{3}}\theta }\) = b

Squaring on both sides

=> b² = \(\frac{sin^{\frac{4}{3}}\theta }{cos^{\frac{2}{3}}\theta }\)

Now, \(a^{2}b^{2}(a^{2}\;+\;b^{2})\)

=> \(\frac{cos^{\frac{4}{3}}\theta }{sin^{\frac{2}{3}}\theta }\) \(\times\) \(\frac{sin^{\frac{4}{3}}\theta }{cos^{\frac{2}{3}}\theta }\)( \(\frac{cos^{\frac{4}{3}}\theta }{sin^{\frac{2}{3}}\theta }\) + \(\frac{sin^{\frac{4}{3}}\theta }{cos^{\frac{2}{3}}\theta }\))

=> \(cos^{\frac{2}{3}}\theta\) \(sin^{\frac{2}{3}}\theta\)( \(\frac{1}{\(cos^{\frac{2}{3}}\theta\)\(cos^{\frac{2}{3}}\theta\]”> \(sin^{\frac{2}{3}}\theta\)}\)

= 1

Hence, L.H.S = R.H.S

Q78. If acos³ \(\theta\) + 3acos\(\theta\) \(sin^{2}\theta\) = m, a\(sin^{3}\theta\) + 3a\(cos^{2}\theta\) \(sin\theta\) = n, prove that \((m\;+\;n)^{\frac{2}{3}}\) + \((m\;-\;n)^{\frac{2}{3}}\) = 2\((a)^{\frac{2}{3}}\)

Ans:

Given, \((m\;+\;n)^{\frac{2}{3}}\) + \((m\;-\;n)^{\frac{2}{3}}\)

Substitute the values of m and n in the above equation

=> (\(( acos³ \(\theta\)3 \(\theta\]”> + 3acos\(\theta\) \(sin^{2}\theta\)\;+\; a\(sin^{3}\theta\) + 3a\(cos^{2}\theta\) \(sin\theta\))^{\frac{2}{3}}\) + (\(( acos³ \(\theta\)3 \(\theta\]”> + 3acos\(\theta\) \(sin^{2}\theta\)\;-\; a\(sin^{3}\theta\) – 3a\(cos^{2}\theta\) \(sin\theta\))^{\frac{2}{3}}\)

=> \((a)^{\frac{2}{3}}\)( (\(( cos³ \(\theta\)3 \(\theta\]”> + 3cos\(\theta\) \(sin^{2}\theta\)\;+\; \(sin^{3}\theta\) + 3\(cos^{2}\theta\) \(sin\theta\))^{\frac{2}{3}}\) + \((a)^{\frac{2}{3}}\)( (\(( cos³ \(\theta\)3 \(\theta\]”> + 3cos\(\theta\) \(sin^{2}\theta\)\;-\; \(sin^{3}\theta\) – 3\(cos^{2}\theta\) \(sin\theta\))^{\frac{2}{3}}\)

=> \((a)^{\frac{2}{3}}\) \(((cos\theta\;+\;sin\theta )^{3})^{\frac{2}{3}}\) + \((a)^{\frac{2}{3}}\) \(((cos\theta\;-\;sin\theta )^{3})^{\frac{2}{3}}\)

=> \((a)^{\frac{2}{3}}\)[ \((cos\theta\;+\;sin\theta )^{2}\)] + \((a)^{\frac{2}{3}}\)[ \((cos\theta\;-\;sin\theta )^{2}\)]

=> \((a)^{\frac{2}{3}}\)( \((cos^{2}\theta \;+\;sin^{2}\theta \;+\;2sin\theta cos\theta )\)) + \((a)^{\frac{2}{3}}\)( \((cos^{2}\theta \;+\;sin^{2}\theta \;-\;2sin\theta cos\theta )\))

=>  \((a)^{\frac{2}{3}}\)[ 1 + \(2sin\theta cos\theta\)] + \((a)^{\frac{2}{3}}\)[ 1 – \(2sin\theta cos\theta\)]

=> \((a)^{\frac{2}{3}}\)[1 + \(2sin\theta cos\theta\)] + 1 – \(2sin\theta cos\theta\)]

=> \((a)^{\frac{2}{3}}\)(1 + 1)

=> 2\((a)^{\frac{2}{3}}\)

Hence, L.H.S = R.H.S

Q79) \(If\;x=acos^{3}\Theta,\;y=bsin^{3}\Theta,\;prove\;that\;(\frac{x}{a})^{\frac{2}{3}}+(\frac{y}{b})^{\frac{2}{3}}=1\)

Ans:

\(x=acos^{3}\Theta:\;y=bsin^{3}\Theta\)

\(\frac{x}{a}=cos^{3}\Theta:\;\frac{y}{b}=sin^{3}\Theta\)

L.H.S = \([\frac{x}{a}]^{\frac{2}{3}}+[\frac{y}{b}]^{\frac{2}{3}}\)

\(=[cos^{3}\Theta]^{\frac{2}{3}}+[sin^{3}\Theta]^{\frac{2}{3}}\)

\(=cos^{2}\Theta+sin^{2}\Theta\;\;\;\;\;\;(because\, cos^{2}\Theta+sin^{2}\Theta=1)\)

=1

Hence proved.

Q80) \(If\;acos\Theta+bsin\Theta=m\;and\;asin\Theta-bcos\Theta=n,\;prove\;that\;a^{2}+b^{2}=m^{2}+n^{2}\)

Ans:

 \(R.H.S=m^{2}+n^{2}\)

\(=(acos\Theta+bsin\Theta)^{2}+(asin\Theta-bcos\Theta)^{2}\\ =a^{2}cos^{2}\Theta+b^{2}sin^{2}\Theta+2absin\Theta cos\Theta+a^{2}sin^{2}\Theta+b^{2}cos^{2}\Theta-2absin\Theta cos\Theta\\ =a^{2}cos^{2}\Theta+b^{2}cos^{2}\Theta+b^{2}sin^{2}\Theta+a^{2}sin^{2}\Theta\\ =a^{2}(sin^{2}\Theta+cos^{2}\Theta)+b^{2}(sin^{2}\Theta+cos^{2}\Theta)\\ =a^{2}+b^{2}\;\;\;\;\;\;\;[because\, sin^{2}\Theta+cos^{2}\Theta=1]\)

= m² + n²

Q81: \(If\;cosA+cos^{2}A=1,\;prove\;that\;sin^{2}A+sin^{4}A=1\)

Ans:

Given- cos A + cos² A = 1

We have to prove sin² A + sin⁴ A = 1

Now, cos A + cos² A = 1

cos A = 1- cos² A

cos A = sin² A

sin² A = cos A

Therefore, we have sin² A + sin⁴ A = cos A + (cos A)² = cos A + cos² A =1

Hence proved.

Q82: \(If\;cos\Theta+cos^{2}\Theta=1,\;prove\;that\;sin^{12}\Theta+3sin^{10}\Theta+3sin^{8}\Theta+sin^{6}\Theta+2sin^{4}\Theta+2sin^{2}\Theta-2=1\)

Ans:

\(cos\Theta+cos^{2}\Theta=1\)

\(cos\Theta=1-cos^{2}\Theta\)

\(cos\Theta=sin^{2}\Theta\)……(i)

\(Now,\;sin^{12}\Theta+3sin^{10}\Theta+3sin^{8}\Theta+sin^{6}\Theta+2sin^{4}\Theta+2sin^{2}\Theta-2\)

\(=(sin^{4}\Theta)^{3}+3sin^{4}\Theta.sin^{2}\Theta(sin^{4}\Theta+sin^{2}\Theta)+(sin^{2}\Theta)^{3}+2(sin^{2}\Theta)^{2}+2sin^{2}\Theta-2\)

\(Using\;(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)\;and\;also\;from\;(i)\;cos\Theta=sin^{2}\Theta\)

\((sin^{4}\Theta+sin^{2}\Theta)^{3}+2cos^{2}\Theta+2cos\Theta-2\)

\(((sin^{2}\Theta)^{2}+sin^{2}\Theta)^{3}+2cos^{2}\Theta+2cos\Theta-2\)

\((cos^{2}\Theta+sin^{2}\Theta)^{3}+2cos^{2}\Theta+2cos\Theta-2\)

\(1+2cos^{2}\Theta+2sin^{2}\Theta-2\;\;\;\;\;[because\, sin^{2}\Theta+cos^{2}\Theta=1]\)

\(1+2(cos^{2}\Theta+sin^{2}\Theta)-2\)

\(1+2(1)-2\)

\(=1\)

L.H.S = R.H.S

Hence proved.

Q83: Given that: \((1+cos\alpha)(1+cos\beta)(1+cos\gamma)=(1-cos\alpha)(1-cos\beta)(1-cos\gamma)\). Show that one of the values of each member of this equality is \(sin\alpha\;sin\beta\;sin\gamma\).

Ans:

We know that \(1+cos\Theta=1+cos^{2}\frac{\Theta}{2}-sin^{2}\frac{\Theta}{2}=2cos^{2}\frac{\Theta}{2}\)

\(\Rightarrow 2cos^{2}\frac{\alpha}{2}.2cos^{2}\frac{\beta}{2}.2cos^{2}\frac{\gamma}{2}…..(i)\)

\(Multiply\;(i)\;with\;sin\alpha\;sin\beta\;sin\gamma\;and\;divide\;it\;with\;same\;we\;get\)

\(\frac{8cos^{2}\frac{\alpha}{2}.cos^{2}\frac{\beta}{2}.cos^{2}\frac{\gamma}{2}}{sin\alpha.sin\beta.sin\gamma}\times sin\alpha.sin\beta.sin\gamma\)

\(\Rightarrow \frac{2cos^{2}\frac{\alpha}{2}.\;cos^{2}\frac{\beta}{2}.\;cos^{2}\frac{\gamma}{2}\times sin\alpha.\;sin\beta.\;sin\gamma}{sin\frac{\alpha}{2}.\;sin\frac{\beta}{2}.\;sin\frac{\gamma}{2}}\)

\(sin\alpha.\;sin\beta.\;sin\gamma\times cot\frac{\alpha}{2}.\;cot\frac{\beta}{2}.\;cot\frac{\gamma}{2}\)

\(RHS\;(1-cos\alpha)(1-cos\beta)(1-cos\gamma)\)

We know that \(1-cos\Theta=1-cos^{2}\frac{\Theta}{2}+sin^{2}\frac{\Theta}{2}=2sin^{2}\frac{\Theta}{2}\)

\(\Rightarrow 2.sin^{2}\frac{\alpha}{2}\;2.sin^{2}\frac{\beta}{2}\;2.sin^{2}\frac{\gamma}{2}\)

\(Multiply\;(i)\;with\;sin\alpha\;sin\beta\;sin\gamma\;and\;divide\;it\;with\;same\;we\;get\)

\(\frac{8sin^{2}\frac{\alpha}{2}.sin^{2}\frac{\beta}{2}.sin^{2}\frac{\gamma}{2}}{sin\alpha.sin\beta.sin\gamma}\times sin\alpha.sin\beta.sin\gamma\)

\(\Rightarrow \frac{8sin^{2}\frac{\alpha}{2}.\;sin^{2}\frac{\beta}{2}.\;sin^{2}\frac{\gamma}{2}\times sin\alpha.\;sin\beta.\;sin\gamma}{2sin\frac{\alpha}{2}cos\frac{\alpha}{2}.\;2sin\frac{\beta}{2}cos\frac{\beta}{2}.\;2sin\frac{\gamma}{2}cos\frac{\gamma}{2}}\)

\(\Rightarrow sin\alpha.\;sin\beta.\;sin\gamma\times tan\frac{\alpha}{2}.\;tan\frac{\beta}{2}.\;tan\frac{\gamma}{2}\)

Hence \(sin\alpha\;sin\beta\;sin\gamma\) is the member of equality.

Q84: If \(sin\Theta+cos\Theta=x,\;prove\;that\;sin^{6}\Theta+cos^{6}\Theta=\frac{4-3(x^{2}-1)^{2}}{4}\).

Ans:

\(sin\Theta+cos\Theta=x\)

Squaring on both sides

\((sin\Theta+cos\Theta)^{2}=x^{2}\)

\(\Rightarrow sin\Theta^{2}+cos\Theta^{2}+2sin\Theta cos\Theta=x^{2}\)

∴ \( sin\Theta cos\Theta=\frac{x^{2}-1}{2}……(i)\)

\(We\;know\;that\;sin^{2}\Theta+cos^{2}\Theta=1\)

Cubing on both sides

\((sin^{2}\Theta+cos^{2}\Theta)^{3}=1^{3}\)

\(sin^{6}\Theta+cos^{6}\Theta+3sin^{2}\Theta cos^{2}\Theta(sin^{2}\Theta+cos^{2}\Theta)=1\)

\(\Rightarrow sin^{6}\Theta+cos^{6}\Theta=1-3sin^{2}\Theta cos^{2}\Theta\)

\(\Rightarrow sin^{6}\Theta+cos^{6}\Theta=1-\frac{3(x^{2}-1)^{2}}{4}\)

∴\( sin^{6}\Theta+cos^{6}\Theta=\frac{4-3(x^{2}-1)^{2}}{4}\)

Q85. If x = \(asec\theta cos\phi\), y =\(bsec\theta sin\phi\) and z= \(c \;tan\phi\), show that \(\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}\;-\;\frac{z^{2}}{c^{2}}\) = 1

Ans:

Given, x = \(asec\theta cos\phi\)

y =\(bsec\theta sin\phi\)

z= \(ctan\phi\)

squaring x,y,z on the sides

\(x^{2}\) = \(a^{2}sec^{2}\theta cos^{2}\phi\)

\(\frac{x^{2}}{a^{2}}\) = \(sec^{2}\theta cos^{2}\phi\)    — 1

\(y^{2}\) = \(b^{2}sec^{2}\theta sin^{2}\phi\)

\(\frac{y^{2}}{b^{2}}\) = \(sec^{2}\theta sin^{2}\phi\)    — 2

\(z^{2}\) = \(c^{2}tan^{2}\phi\)

\(\frac{z^{2}}{c^{2}}\) = \(tan^{2}\phi\)              —— 3

Substitute  eq 1,2,3 in   \(\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}\;-\;\frac{z^{2}}{c^{2}}\)

=> \(\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}\;-\;\frac{z^{2}}{c^{2}}\)

=> \(sec^{2}\theta cos^{2}\phi\) + \(sec^{2}\theta sin^{2}\phi\) – \(tan^{2}\phi\)

=> \(sec^{2}\theta (cos^{2}\phi \;+\;sin^{2}\phi )\) – \(tan^{2}\phi\)

We know that, \(cos^{2}\phi \;+\;sin^{2}\phi\) = 1

=> \(sec^{2}\theta\)(1) – \(tan^{2}\phi\)

And, \(sec^{2}\theta \;-\;tan^{2}\theta\) = 1

=> 1

Hence, L.H.S= R.H.S

Q86. If \(sin\theta \;+\;2cos\theta\) prove that \(2sin\theta \;-\;cos\theta\) = 2

Ans:

Given, \(sin\theta \;+\;2cos\theta\) =1

Squaring on both sides

=> \((sin\theta \;+\;2cos\theta)^{2}\) = 1²

=> \(sin^{2}\theta \;+\;4cos^{2}\theta\;+\;4sin\theta cos\theta\) = 1

=>  \(4cos^{2}\theta\;+\;4sin\theta cos\theta\) = 1 – \(sin^{2}\theta\)

Here, 1 – \(sin^{2}\theta\) = \(cos^{2}\theta\)

=> \(4cos^{2}\theta\;+\;4sin\theta cos\theta\) – \(cos^{2}\theta\) = 0

=> \(3cos^{2}\theta\;+\;4sin\theta cos\theta\) =0           —- 1

We have, \(2sin\theta \;-\;cos\theta\) = 2

Squaring L.H.S

\((2sin\theta\;-\;cos\theta )^{2}\) = \(4sin^{2}\theta \;+\;cos^{2}\theta\;-\;4sin\theta cos\theta\)

Here, \(4sin\theta cos\theta\) = \(3cos^{2}\theta\)

= \(4sin^{2}\theta \;+\;cos^{2}\theta \;+\;3cos^{2}\theta\)

= \(4sin^{2}\theta \;+\;4cos^{2}\theta\)

= \(4(sin^{2}\theta \;+\;cos^{2}\theta)\)

=  4(1)

= 4

\((2sin\theta\;-\;cos\theta )^{2}\) = 4

=> \(2sin\theta \;-\;cos\theta\) = 2

Hence proved