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
Â
Â
Q3:Â \(tan^{2}\theta\;cos^{2}\theta\)
A3: We know ,
\(sin^{2}\theta + cos^{2}\theta=1\)
So,
\(tan^{2}\theta\;cos^{2}\theta\)
\(= (\frac{sin\theta}{cos\theta}\times cos\theta)^{2}\)
\(= (sin\theta)^{2}\)
\(= sin^{2}\theta\)
\(1-cos^{2}\theta\)
Â
Â
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
Â
Â
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
Â
Â
Q6:Â \(tan\theta+ \frac{1}{tan\theta} =sec\theta\;cosec\theta\)
A6: We know that,
\((sec^{2}\theta-tan^{2}\theta)= 1\)
So,
\(tan\theta+ \frac{1}{tan\theta} =\frac{tan^{2}\theta+1}{tan\theta}\)
\(=\frac{sec^{2}\theta}{tan\theta}\)
\(=sec\theta\frac{sec\theta}{tan\theta}\)
\(=sec\theta\frac{\frac{1}{cos\theta}}{\frac{sin\theta}{cos\theta}}\)
\(=sec\theta\frac{1}{sin\theta}\)
\(=sec\thetacosec\theta\)
Â
Â
Q7: \(\frac{cos\theta}{1-sin\theta}=\frac{1+sin\theta}{cos\theta}\)
A7:Â We know ,
\(sin^{2}\theta + cos^{2}\theta=1\)
So, Multiplying both numerator and denominator by \({(1+sin\theta)}\)
\(\frac{cos\theta}{1-sin\theta}=\frac{cos\theta(1+sin\theta)}{(1-sin\theta)(1+sin\theta)}\)
\(=\frac{cos\theta(1+sin\theta)}{(1-sin^{2}\theta)}\)
\(=\frac{cos\theta(1+sin\theta)}{cos^{2}\theta}\)
\(=\frac{(1+sin\theta)}{cos\theta}\)
Â
Â
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 \({(1-sin\theta)}\)
\(\frac{cos\theta}{1+sin\theta}=\frac{cos\theta(1-sin\theta)}{(1+sin\theta)(1-sin\theta)}\)
\(=\frac{cos\theta(1-sin\theta)}{(1-sin^{2}\theta)}\)
\(=\frac{cos\theta(1-sin\theta)}{(cos^{2}\theta)}\)
\(=\frac{(1-sin\theta)}{cos\theta}\)
\(=\frac{(1-sin\theta)}{cos\theta}\)
Â
Â
Q 9:cos2A +\(\frac{1}{1+cot^{2}A}\)
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
Â
Â
Q10: \(sinA^{2}+\frac{1}{1+tan^{2}A}=1\)
A10:Â Â Â Â 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
Â
Â
Q11: \(\sqrt{\frac{1-cos\theta }{1+cos\theta}}=cosec\theta-cot\theta\)
A11:Â We know ,
\(sin^{2}\theta + cos^{2}\theta=1\)
Multiplying both numerator and denominator by \({(1-cos\theta)}\)
\(\sqrt{\frac{1-cos\theta }{1+cos\theta}}=\sqrt{\frac{(1-cos\theta)(1-cos\theta) }{(1+cos\theta)(1-cos\theta)}}\)
\(=\sqrt{\frac{(1-cos\theta)^{2}}{1-cos^{2}\theta}}\)
\(=\sqrt{\frac{(1-cos\theta)^{2}}{sin^{2}\theta}}\)
\(={\frac{(1-cos\theta)}{sin\theta}}\)
\(=\frac{1}{sin\theta}-\frac{cos\theta}{sin\theta}\)
\(=\frac{1}{sin\theta}-\frac{cos\theta}{sin\theta}[/latex\]
Â
Â
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 \({(1+cos\theta)}\)
\(=\frac{(1-cos^{2}\theta)}{(1+cos\theta)(sin\theta)}\)
\(=\frac{(sin^{2}\theta)}{(1+cos\theta)(sin\theta)}\)
\(=\frac{(sin\theta)}{(1+cos\theta)}\)
Â
Â
Q13. \(\frac{sin\theta}{1-cos\theta}\)
Ans:
Given, L.H.S = \(\frac{sin\theta}{1-cos\theta}\)
Rationalize both nr and dr with 1+cos \(\theta\)
= \(\frac{sin\theta}{1-cos\theta}\)
We know that, (a-b)(a+b) = a2 – b2
=> \(\frac{sin\theta(1+cos\theta)}{1-cos^{2}\theta}\)
Here, (1-cos2 \(\theta\)
=> \(\frac{sin\theta\;+\;(sin\theta*cos\theta)}{sin^{2}\theta}\)
=> \(\frac{sin\theta }{sin^{2}\theta}\)
=> \(\frac{1}{sin\theta}\)
=> cosec \(\theta\)
Hence, L.H.S = R.H.S
Â
Â
Q14. \(\frac{1\;-\;sin\theta}{1\;+\;sin\theta}\)
Ans:
Given, L.H.S = \(\frac{1\;-\;sin\theta}{1\;+\;sin\theta}\)
Rationalize with nr and dr with 1 – sin \(\theta\)
=> \(\frac{1\;-\;sin\theta}{1\;+\;sin\theta}\)
Here, (1-sin \(\theta\)
=> \(\frac{(1\;-\;sin\theta)^{2}}{cos^{2}\theta}\)
=> \((\frac{1\;-\;sin\theta}{cos\theta})^{2}\)
=> \((\frac{1}{cos\theta}\;-\;\frac{sin\theta}{cos\theta})^{2}\)
=> \((sec\theta\;-\;tan\theta)^{2}\)
Hence, L.H.S = R.H.S
Â
Â
Q15. \(\frac{(1\;+\;cot^{2}\theta)tan\theta}{sec^{2}\theta}\)
Ans:
Given, L.H.S = \(\frac{(1\;+\;cot^{2}\theta)tan\theta}{sec^{2}\theta}\)
Here, 1 + cot2 \(\theta\)
=> \(\frac{cosec^{2}\theta*tan\theta}{sec^{2}\theta}\)
=> \(\frac{1}{sin^{2}\theta}\)
=> \(\frac{cos\theta}{sin\theta}\)
=> cot \(\theta\)
Hence, L.H.S = R.H.S
Â
Â
Q16. \(tan^{2}\theta\;-\;sin^{2}\theta\)
Ans:
Given, L.H.S = \(tan^{2}\theta\;-\;sin^{2}\theta\)
Here, tan2 \(\theta\)
=> \(\frac{sin^{2}\theta}{cos^{2}\theta}\)
=> \(sin^{2}\theta\)
=> \(sin^{2}\theta\)
=> \(\frac{sin^{2}\theta}{cos^{2}\theta}\)
=> \(tan^{2}\theta\;*\;sin^{2}\theta\)
Hence, L.H.S = R.H.S
Â
Â
Q17. (cosec \(\theta\)
Ans:
Given, L.H.S = (cosec \(\theta\)
Here, (a + b)(a – b) = a2 – b2
cosec2 \(\theta\)
=>Â 1 + cot2 \(\theta\)
=>Â 1 + cot2 \(\theta\)
=> cot2 \(\theta\)
Hence, L.H.S = R.H.S
Â
Â
Q18. (\(sec\theta\)
Ans:
Given, L.H.S = (sec \(\theta\)
Here, (a + b)(a – b) = a2 – b2
sec2\(\theta\)
=> 1 + tan2 \(\theta\)
=> 1 + tan2 \(\theta\)
=>Â tan2Â \(\theta\)
Hence, L.H.S = R.H.S
Â
Â
Q19. secA(1 – sinA)(secA + tanA) = 1
Ans:
Given, L.H.S = secA(1 – sinA)(secA + tanA)
Here, secA = \(\frac{1}{cosA}\)
=> \(\frac{1}{cosA}\)
=> \(\frac{cos2A}{cos2A}\)
Substitute the above values in L.H.S
=> (\(\frac{1}{sinA}\)
=> (\(\frac{1\;-sin^{2}A}{sinA}\)
\(\frac{sin^{2}A\;+\;cos^{2}A}{sinA\;*\;cosA}\)
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{sin^{2}A\;*\;cos^{2}A\;*\;1}{sin^{2}A\;*\;cos^{2}A}\]”>
=> 1
Hence, L.H.S = R.H.S
Â
Â
Q21. (1 + \(tan^{2}\theta\)
Ans:
Given, L.H.S = (1 + \(tan^{2}\theta\)
We know that,
Sin2 \(\theta\)
And sec2 \(\theta\)
So,
(1 + \(tan^{2}\theta\)
= (1 + \(tan^{2}\theta\)
= \(sec^{2}\theta\;*\;tan^{2}\theta\)
= \((\frac{1}{cos^{2}\theta})\;*\;cos^{2}\theta\)
= 1
hence, L.H.S = R.H.S
Â
Â
Q22. \((sin^{2}A\;*\;cot^{2}A)\;+\;(cos^{2}A;*\;tan^{2}A)\)
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,
\((sin^{2}\A\;*\;cot^{2}A)\;+\;(cos^{2}A\;*\;tan^{2}A)\)
= cos2A + sin2A
= 1
Hence , L.H.S = R.H.S
Â
Â
Q23:
- cot \(\theta\)
– tan \(\theta\) = \(\frac{2cos^{2}\theta\;-\;1}{sin\theta*cos\theta}\)
Ans:
Give, L.H.S = cot \(\theta\)
Here, Sin2 \(\theta\)
So,
=> cot \(\theta\)
= \(\frac{cos^{2}\theta\;-\;sin^{2}\theta}{sin\theta\;*\;cos\theta }\)
= \(\frac{cos^{2}\theta\;-\;(1\;-\;cos^{2}\theta)}{sin\theta\;*\;cos\theta }\)
= \(\frac{cos^{2}\theta\;-\;1\;-\;cos^{2}\theta}{sin\theta\;*\;cos\theta }\)
= \((\frac{2cos^{2}\theta\;-\;1}{sin\theta*cos\theta})\)
Hence, L.H.S = R.H.S
Â
Â
- \(tan\theta\;-\;cot\theta\;=\;\)
\((\frac{2sin^{2}\theta\;-\;1}{sin\theta*cos\theta})\)
Sol:
Given, L.H.S = \(tan\theta\;-\;cot\theta
We know that,
Sin2 \(\theta\)
\(tan\theta\;-\;cot\theta = \(\frac{sin\theta }{cos\theta }\)
= \(\frac{sin^{2}\theta\;-\;cos^{2}\theta }{sin\theta cos\theta }\)
= \(\frac{sin^{2}\theta\;-\;(1\;-\;sin^{2}\theta) }{sin\theta cos\theta }\)
= \(\frac{sin^{2}\theta\;-\;1\;+\;sin^{2}\theta }{sin\theta cos\theta }\)
= \((\frac{2sin^{2}\theta\;-\;1}{sin\theta*cos\theta})\)
Hence, L.H.S = R.H.S
Â
Â
Q24. \(\frac{cos^{2}\theta }{sin\theta }\;-\;cosec\theta\;+\;sin\theta\)
Ans:
Given, L.H.S \(\frac{cos^{2}\theta }{sin\theta }\;-\;cosec\theta\;+\;sin\theta\)
We know that,
Sin2 \(\theta\)
So,
\(\frac{cos^{2}\theta }{sin\theta }\;-\;cosec\theta\;+\;sin\theta\)
= \((\frac{cos^{2}\theta }{sin\theta }\;-\;\frac{1}{sin\theta })\;+\;sin\theta\)
= \((\frac{cos^{2}\theta\;-\;1 }{sin\theta })\;+\;sin\theta\)
= \((\frac{-sin^{2}\theta }{sin\theta })\;+\;sin\theta\)
= \(-sin\theta \;+\;sin\theta\)
= 0
Hence, L.H.S = R.H.S
Â
Â
Q 25 . \(\frac{1}{1 + sin\;A}\)
Ans:
LHS = \(\frac{1}{1 + sin\;A}\)
\(\frac{\left ( 1-sinA \right )+\left ( 1+sinA \right )}{\left ( 1+sinA \right )\left ( 1-sinA \right )}\)
\(\frac{1-sinA+1+sinA}{1-sin^{2}A}\)
\(\Rightarrow \frac{2}{1-sin^{2}\;A}\)
\(\Rightarrow \frac{2}{cos^{2}A}\)
\(\Rightarrow 2sec^{2}A\)
\(\therefore\)
Â
Â
Q 26 . \(\frac{1+sin \theta }{cos \theta}+\frac{cos \theta}{1 + sin \theta} = 2sec \theta\)
Ans:
LHS = \(\frac{1+sin \theta }{cos \theta}+\frac{cos \theta}{1 + sin \theta}\)
= \(\frac{\left (1+sin \theta \right )^{2}+cos^{2}\theta }{cos \theta\left ( 1 + sin \theta \right )}\)
= \(\frac{1+sin^{2}\theta +2sin \theta +cos^{2}\theta }{cos \theta\left ( 1 + sin \theta \right )}\)
\(\Rightarrow \frac{2\left ( 1+sin \theta \right ) }{cos \theta\left ( 1 + sin \theta \right )}\)
\(\therefore\)
Â
Â
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 = \(\frac{\left ( 1+sin\theta \right )^{2}+\left ( 1-sin\theta \right )^{2}}{2cos^{2}\theta }\\=\frac{\left ( 1+2sin\theta+sin^{2} \theta \right )+\left ( 1-2sin\theta+sin^{2} \theta \right )}{2cos^{2}\theta}\\=\frac{ 1+2sin\theta+sin^{2} \theta +1-2sin\theta+sin^{2} \theta }{2cos^{2}\theta}\\=\frac{2+2sin^{2}\theta }{2cos^{2}\theta }\\=\frac{2\left ( 1+sin^{2}\theta \right )}{2\left ( 1-sin^{2}\theta \right )}\\=\frac{\left ( 1+sin^{2}\theta \right )}{\left ( 1-sin^{2}\theta \right )}\)
\(\therefore\)
Â
Â
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 = \(\frac{1+tan^{2}\theta }{1+cot^{2}\theta }\)
= \(\frac{sec^{2}\theta }{cosec^{2}\theta }\)
= \(\frac{1}{cos^{2}\theta \cdot 1}sin^{2}\theta\)
= \(tan^{2}\theta\)
\(\therefore\)
Â
Â
Q 29 . \(\frac{1+sec\theta} {sec\theta}\)
Ans :
LHS =Â \(\frac{1+sec\theta} {sec\theta}\)
= \(\frac{1+\frac{1}{cos\theta }}{\frac{1}{cos\theta }}\)
= \(\frac{cos\theta +1}{cos\theta }\cdot cos\theta\)
= \(1 + cos\theta\)
RHS = \(\frac{sin^{2}\theta }{1-cos\theta }\)
= \(\frac{1 – cos^{2}\theta }{1-cos\theta }\)
= \(\frac{\left ( 1-cos\theta \right )\left ( 1+cos\theta \right )}{1-cos\theta }\)
= \(1 + cos\theta\)
\(\therefore\)
Â
Â
Q 30 . \(\frac{tan\theta }{1-cot\theta }+\frac{cot\theta }{1-tan\theta}\)
Ans:
LHS = \(\frac{tan\theta }{1-\frac{1}{tan\theta }}+\frac{cot\theta }{1-tan\theta }\)
= \(\frac{tan^{2}\theta }{tan\theta -1}+\frac{cot\theta }{1-tan\theta }\)
= \(\frac{1 }{1-tan\theta }\left [ \frac{1}{tan\theta } -tan^{2}\theta \right ]\)
= \(\frac{1 }{1-tan\theta }\left [ \frac{1-tan^{3}\theta}{tan\theta } \right ]\)
= \(\frac{1 }{1-tan\theta }\frac{\left ( 1-tan\theta \right )\left ( 1+tan\theta +tan^{2}\theta \right )}{tan\theta}\)
= \(\frac{ 1+tan\theta +tan^{2}\theta }{tan\theta}\)
= \(\frac{1}{tan\theta }+\frac{tan\theta}{tan\theta}+\frac{tan^{2}\theta}{tan\theta}\)
=Â \(1 + tan\theta +cot\theta\)
\(\therefore\)
Â
Â
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
\(\left (sec^{2}\theta -tan^{2}\theta \right )^{3} = 1\)
\(sec^{6}\theta -tan^{6}\theta – 3 sec^{2}\theta tan^{2}\theta \left ( sec^{2}\theta -tan^{2}\theta \right ) = 1\)
\(sec^{6}\theta -tan^{6}\theta – 3 sec^{2}\theta tan^{2}\theta = 1\)
\(\Rightarrow sec^{6}\theta = tan^{6}\theta+3 sec^{2}\theta tan^{2}\theta +1\)
Hence proved.
Â
Q 32 . \(cosec^{6}\theta = cot^{6}\theta+3 cot^{2} \theta cosec^{2}\theta +1\)
Ans :
We know that \(cosec^{2}\theta –Â cot^{2}\theta = 1\)
Cubing both sides
\(\left (cosec^{2}\theta -cot^{2}\theta \right )^{3} = 1\)
\(cosec^{6}\theta -cot^{6}\theta – 3 cosec^{2}\theta cot^{2}\theta \left ( cosec^{2}\theta -cot^{2}\theta \right ) = 1\)
\(cosec^{6}\theta -cot^{6}\theta – 3 cosec^{2}\theta cot^{2}\theta = 1\)
\(\Rightarrow cosec^{6}\theta = cot^{6}\theta+3 cosec^{2}\theta cot^{2}\theta +1\)
Hence proved.
Â
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 = \(\frac{sec^{2}\theta \cdot cot\theta }{cosec^{2}\theta }\)
= \(\frac{1\cdot sin^{2}\theta }{cos^{2}\theta }\cdot \frac{cos\theta }{sin\theta }\)
\(\Rightarrow \frac{sin\theta }{cos\theta }= tan\theta\)
\(\therefore\)
Â
Q 34 . \(\frac{ 1+cosA}{sin ^{2} A}\)
Ans:
We know that \(sin^{2} A+cos^{2}A\)
\(sin^{2}A=1-cos^{2}A\)
\(\Rightarrow sin^{2}A=\left (1-cosA \right )\left ( 1+cosA \right )\)
\(\Rightarrow\;LHS = \frac{\left ( 1+cosA \right )}{\left ( 1-cosA \right )\left ( 1+cosA \right )}\)
= \(\Rightarrow\;LHS = \frac{1}{\left ( 1-cosA \right )}\)
\(\therefore\)
Â
Â
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
\(\frac{sec A-tanA}{sec A+tanA}\times \frac{sec A+tanA}{sec A+tanA}\)
= \(\frac{sec^{2} A-tan^{2}A}{\left (sec A+tanA \right )^{2}}\)
= \(\frac{1}{\left (sec A+tanA \right )^{2}}\)
= \(\frac{1}{\left (sec ^{2}A+tan^{2}A +2sec A tan A \right )}\)
= \(\frac{1}{\left (\frac{1}{cos ^{2}A}+\frac{sin ^{2}A}{cos ^{2}A} +\frac{2sinA}{cosA} \right )}\)
\(\Rightarrow \frac{cos ^{2}A}{1 +sin^{2}A+2 sin A}\)
= \(\frac{cos ^{2}A}{\left (1 +sinA \right )^{2}}\)
\(\therefore\)
Â
Â
Q 36 . \(\frac{1+cosA}{sinA}\)
Ans:
LHS = \(\frac{1+cosA}{sinA}\)
Multiply both numerator and denominator with (1 – cos A) we get ,
\(\frac{\left (1+cosA \right )\left ( 1-cosA \right )}{sin A\left (1-cosA \right )}\)
= \(\frac{1-cos^{2}A }{sin A\left (1-cosA \right )}\)
= \(\frac{sin^{2}A }{sin A\left (1-cosA \right )}\)
= \(\frac{sinA }{1-cosA }\)
\(\therefore\)
Â
Â
37.
 (i) \(\sqrt{\frac{1+sin A}{1- sin A}}\)
Ans:
To prove,
\(\sqrt{\frac{1+sin A}{1- sin A}}\)
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
Â
(ii) \(\sqrt{\frac{(1-cos A)}{(1+cos A)}}+\sqrt{\frac{(1+cos A)}{(1-cos A)}}\)
Ans:
To prove,
\(\sqrt{\frac{(1-cos A)}{(1+cos A)}}+\sqrt{\frac{(1+cos A)}{(1-cos A)}}\)
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
Â
Â
- Prove that:
(i) \(\sqrt{\frac{(sec \Theta- 1)}{(sec \Theta+1)}}+\sqrt{\frac{(sec \Theta+ 1)}{(sec \Theta-1)}}\)
Ans:
To prove,
= \(\sqrt{\frac{(sec \Theta- 1)}{(sec \Theta+1)}}+\sqrt{\frac{(sec \Theta+ 1)}{(sec \Theta-1)}}\)
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
Â
Â
(ii) \(\sqrt{\frac{( 1+sin\Theta)}{( 1-sin\Theta)}}+\sqrt{\frac{( 1-sin\Theta)}{( 1+sin\Theta)}}\)
Ans:
To prove,
= \(\sqrt{\frac{( 1+sin\Theta)}{( 1-sin\Theta)}}+\sqrt{\frac{( 1-sin\Theta)}{( 1+sin\Theta)}}\)
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
Â
(iii) \( \sqrt{\frac{(1+cos \Theta)}{(1-cos \Theta)}}\)
Ans:
To prove,
\(\sqrt{\frac{(1-cos \Theta)}{(1+cos \Theta)}}+\sqrt{\frac{(1+cos \Theta)}{(1-cos \Theta)}}\)
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) \(\frac{sec\Theta-1}{sec\Theta+1}\)
Ans:
To prove,
\(\frac{sec\Theta-1}{sec\Theta+1}\)
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
Â
Â
- (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
Â
Â
- \(\frac{1 – cos A}{1 + cos A}\)
= (cot A – cosec A)2
Ans:
To prove,
\(\frac{1- cos A}{1 + cos A}\)
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
Â
- \(\frac{1}{sec A – 1}+\frac{1}{sec A+1} = 2 cosec A cot A\)
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
Â
- \(\frac{cos A}{1-tan A}+\frac{sin A}{1-cot A}\)
= sin A + cos A
Ans:
To prove,
\(\frac{cos A}{1-tan A}+\frac{sin A}{1-cot A}\)
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
Â
Â
- \(\frac{(cosec A)}{(cosec A-1)}+\frac{(cosec A)}{(cosec A+1)}\)
= 2sec2 A
Ans:
To prove,
\(\frac{(cosec A)}{(cosec A-1)}+\frac{(cosec A)}{(cosec A+1)}\)
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.\(\frac{tan^{2}A}{1+tan^{2}A}+\frac{cot^{2}A}{1+cot^{2}A}\)
Ans:
To prove,
\(\frac{tan^{2}A}{1+tan^{2}A}+\frac{cot^{2}A}{1+cot^{2}A}\)
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
- \(\frac{cot A-cos A}{cot A+cos A}\)
= \(\frac{cosec A-1}{cosec A + 1}\)
Ans:
To prove,
\(\frac{cot A-cos A}{cot A+cos A}\)
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
Â
Â
- \(\frac{1+cos\Theta -sin^{2}\Theta }{sin\Theta (1+cos\Theta)}\)
= cot \(\Theta\)
Ans:
To prove,
\(\frac{1+cos\Theta -sin^{2}\Theta }{sin\Theta (1+cos\Theta)}\)
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.
Â
(i) \(\frac{1+cos\Theta +sin\Theta }{1+cos\Theta -sin\Theta }\)
Ans:
To prove,
\(\frac{1+cos\Theta +sin\Theta }{1+cos\Theta -sin\Theta }\)
Dividing the numerator and denominator with \(cos\Theta\)
Considering LHS, we get,
= \(\frac{\frac{1+cos\Theta +sin\Theta}{cos\Theta }}{\frac{1+cos\Theta -sin\Theta}{cos\Theta }}\)
= \(\frac{sec\Theta+1+tan\Theta }{sec\Theta+1-tan\Theta }\)
= \(\frac{1+sec\Theta+tan\Theta }{1+sec\Theta-tan\Theta }\)
[As we know,
\((sec^{2}\Theta )-(tan^{2}\Theta ) = 1\\ (sec\Theta+tan\Theta)(sec\Theta-tan\Theta ) = 1\\ (sec\Theta+tan\Theta) = \frac{1}{(sec\Theta-tan\Theta)}\)
= \(\frac{\frac{1}{(sec\Theta-tan\Theta)}+1}{1+sec\Theta-tan\Theta}\)
= \(\frac{1+sec\Theta-tan\Theta}{1+sec\Theta-tan\Theta}\times \frac{1}{sec\Theta-tan\Theta}\)
= \(sec\Theta+tan\Theta\)
= \(\frac{1+sin\Theta }{cos\Theta }\)
Therefore, LHS = RHS
Hence proved
(ii)Â \(\frac{sin\Theta-cos\Theta+1 }{sin\Theta+cos\Theta-1}\)
Ans:
To prove,
\(\frac{sin\Theta-cos\Theta+1 }{sin\Theta+cos\Theta-1}\)
Considering LHS, we get,
\(\frac{sin\Theta-cos\Theta+1 }{sin\Theta+cos\Theta-1}\)
Dividing the numerator and denominator with \(cos\Theta\)
= \(\frac{tan\Theta+sec\Theta-1 }{tan\Theta-sec\Theta+1}\)
[As we know, \((sec\Theta+tan\Theta) = \frac{1}{(sec\Theta-tan\Theta)}\)
= \(\frac{ \frac{1}{(sec\Theta-tan\Theta)}-1}{tan\Theta-sec\Theta+1}\)
= \(\frac{tan\Theta-sec\Theta+1}{tan\Theta-sec\Theta+1}\times \frac{1}{(sec\Theta-tan\Theta)}\)
= \(\frac{1}{(sec\Theta-tan\Theta)}\)
Therefore, LHS = RHS
Hence proved
Â
(iii) \(\frac{cos\Theta-sin\Theta+1 }{cos\Theta+sin\Theta-1}\)
Ans:
To prove,
\(\frac{cos\Theta-sin\Theta+1 }{cos\Theta+sin\Theta-1}\)
Considering LHS, we get,
Dividing the numerator and denominator with \(sin\Theta\)
= \(\frac{\frac{cos\Theta-sin\Theta+1}{sin\Theta } }{\frac{cos\Theta+sin\Theta-1}{sin\Theta}}\)
= \(\frac{cot\Theta+cosec\Theta-1}{cot\Theta-cosec\Theta+1}\)
[As we know, \((cosec^{2}\Theta )-(cot^{2}\Theta ) = 1\\ (cosec\Theta+cot\Theta)(cosec\Theta-cot\Theta ) = 1\\ (cosec\Theta+cot\Theta) = \frac{1}{(cosec\Theta-cot\Theta)}\)
= \(\frac{\frac{1}{(cosec\Theta-cot\Theta)}-1}{cot\Theta-cosec\Theta+1}\)
= \(\frac{cot\Theta-cosec\Theta+1}{cot\Theta-cosec\Theta+1}\times \frac{1}{(cosec\Theta-cot\Theta)}\)
= \(\frac{1}{(cosec\Theta-cot\Theta)}\)
= \(cosec\Theta + cot\Theta\)
Therefore, LHS = RHS
Hence proved
(iv) \((sin\Theta + cos\Theta )(tan\Theta + cot\Theta)\)
Ans:
To prove,
\((sin\Theta + cos\Theta )(tan\Theta + cot\Theta)\)
Considering LHS, we get,
= \((sin\Theta + cos\Theta)(\frac{sin\Theta}{cos\Theta }+\frac{cos\Theta}{sin\Theta})\)
= \((\frac{sin^{2}\Theta}{cos\Theta}+ cos\Theta+sin\Theta+\frac{cos^{2}\Theta}{sin\Theta}\)
= \(sin\Theta(tan\Theta +1) +cos\Theta( \frac{1}{tan\Theta}+1)\)
= \(sin\Theta(tan\Theta +1) +\frac{cos\Theta}{tan\Theta}(tan\Theta +1)\)
= \((sin\Theta+\frac{cos\Theta}{tan\Theta})(tan\Theta +1)\)
= \((\frac{sin^{2}\Theta+cos^{2}\Theta}{sin\Theta})(tan\Theta +1)\)
= \((\frac{1}{sin\Theta})(tan\Theta +1)\)
= \((\frac{sin\Theta +cos\Theta }{cos\Thetasin\Theta })\)
= \(sec\Theta+cosec\Theta\)
Therefore, LHS = RHS
Hence proved
Â
Â
- \frac{tanA}{1+secA}-\frac{tanA}{1-secA}= 2 cosec A
Ans:
To prove,
\frac{tanA}{1+secA}-\frac{tanA}{1-secA}= 2 cosec A
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
Therefore, LHS = RHS
Hence proved
Q51: \(1+\frac{cot^{2}\Theta}{1+cosec\Theta}=cosec\Theta\)
Ans:
\(1+\frac{cosec^{2}\Theta-1}{1+cosec\Theta}\;\;\;\;[\because cot^{2}\Theta=cosec^{2}\Theta-1]\)
\(1+\frac{(cosec\Theta-1)(cosec\Theta+1)}{1+cosec\Theta}\)
\(=1+cosec\Theta-1\;\;\;\;\;[\because (a+b)(a-b)=a^{2}-b^{2})]\)
\(=cosec\Theta\)
Therefore, 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
Â
Â
Q53) \((1+tan^{2}A)+(1+\frac{1}{tan^{2}A})=\frac{1}{sin^{2}A-sin^{4}A}\)
Ans:
LHS = \((1+\frac{sin^{2}A}{cos^{2}A})+(1+\frac{cos^{2}A}{sin^{2}A})\)
=>\(\frac{cos^{2}A+sin^{2}A}{cos^{2}A}+\frac{sin^{2}A+cos^{2}A}{sin^{2}A}\)
=>\(\frac{1}{cos^{2}A}+\frac{1}{sin^{2}A}\;\;\;\;\;\;[\because sin^{2}A+cos^{2}A=1]\)
=>\(\frac{sin^{2}A+cos^{2}A}{sin^{2}Acos^{2}A}=\frac{1}{sin^{2}A(1-sin^{2}A)}\;\;\;\;\;\;[\because cos^{2}A=1-sin^{2}A]\)
=>\(\frac{1}{sin^{2}A-sin^{4}A}\)
Therefore, LHS = RHS.
Hence Proved.
Â
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.
Â
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.
Â
Â
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 = \(asec\theta+btan\theta\)
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.
Â
Â
Q59) If \(3sin\theta+5cos\theta=5\)
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.
Â
Q60) If \(cosec\theta+cot\theta\)
Ans:
LHS = mn
= \((cosec\theta+cot\theta)(cosec\theta-cot\theta)\)
= \(cosec^{2}\theta-cot^{2}\theta\)
= 1
= RHS
Hence Proved.
Â
Q 62 . If \(T_{n}=sin^{n}\theta +cos_{n}\theta\)
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 = \(\frac{\left (sin^{5}\theta +cos^{5}\theta \right ) – \left (sin^{7}\theta +cos^{7}\theta \right ) }{sin^{3}\theta +cos^{3]\theta }\)
= \(\frac{sin^{5}\theta \left ( 1- sin^{2}\theta \right )+cos^{5}\theta \left ( 1- cos^{2}\theta \right ) }{sin^{3]\theta +cos^{3}\theta }\)
= \(\frac{sin^{5}\theta \times cos^{2}\theta +cos^{5}\theta \times sin^{2}\theta }{sin^{3}\theta +cos^{3}\theta }\)
= \(\frac{sin^{2}\theta cos^{2}\theta \left ( sin^{3}\theta +cos^{3}\theta \right ) }{sin\theta +cos\theta }\)
= \(sin^{2}\theta cos^{2}\theta\)
\(\therefore\)
Â
Â
Q 63 . \(\left ( tan\theta +\frac{1}{cos\theta } \right )^{2}+\left ( tan\theta -\frac{1}{cos\theta } \right )^{2}\)
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
\(\therefore\)
Â
Â
Q 64 . \(\left (\frac{1}{sec^{2}\theta -cos^{2}\theta}+\frac{1}{cosec^{2}\theta -sin^{2}} \right )sin^{2}\theta cos^{2}\theta\)
Ans:
\(\left [ \frac{1}{\frac{1}{cos^{2}\theta }-cos^{2}\theta } +\frac{1}{\frac{1}{sin^{2}\theta }-sin^{2}\theta}\right ]sin^{2}\theta cos^{2}\)
= \(\left [ \frac{1}{\frac{1-cos^{4}\theta}{cos^{2}\theta } } +\frac{1}{\frac{1-sin^{4}\theta }{sin^{2}\theta}}\right ]sin^{2}\theta cos^{2}\theta \)
= \(\left [ \frac{cos^{2}\theta }{1-cos^{4}\theta} +\frac{sin^{2}\theta}{1-sin^{4}\theta}\right ]sin^{2}\theta cos^{2}\theta\)
= \(\left [ \frac{cos^{2}\theta }{cos^{2}\theta+sin^{2}\theta-cos^{4}\theta} +\frac{sin^{2}\theta}{cos^{2}\theta+sin^{2}\theta-sin^{4}\theta}\right ]sin^{2}\theta cos^{2}\theta\)
= \(\left [ \frac{cos^{2}\theta }{cos^{2}\theta\left ( 1-cos^{2}\theta \right )+sin^{2}\theta} +\frac{sin^{2}\theta}{cos^{2}\theta+sin^{2}\theta\left ( 1- sin^{2}\theta\right )}\right ]sin^{2}\theta cos^{2}\theta\)
= \(\left [ \frac{cos^{2}\theta }{cos^{2}\theta sin^{2}\theta+sin^{2}\theta} +\frac{sin^{2}\theta}{cos^{2}\theta+sin^{2}\theta cos^{2}\theta}\right ]sin^{2}\theta cos^{2}\theta\)
= \(\left [ \frac{cos^{2}\theta }{sin^{2}\theta \left ( cos^{2}\theta +1 \right )} +\frac{sin^{2}\theta}{cos^{2}\theta\left ( sin^{2}\theta+1 \right )}\right ]sin^{2}\theta cos^{2}\theta\)
= \(\frac{cos^{4}\theta \left ( sin^{2}\theta+1 \right )+sin^{4}\theta\left ( cos^{2}\theta +1 \right )}{sin^{2}\theta cos^{2}\theta \left ( cos^{2}\theta +1 \right )\left (sin^{2}\theta+1 \right ) } sin^{2}\theta cos^{2}\theta\)
= \(\frac{cos^{4}\theta \left ( sin^{2}\theta+1 \right )+sin^{4}\theta\left ( cos^{2}\theta +1 \right )}{ \left ( cos^{2}\theta +1 \right )\left (sin^{2}\theta+1 \right ) }\)
= \(\frac{cos^{4}\theta +cos^{4}\theta sin^{2}\theta +sin^{4}\theta +sin^{4}\theta cos^{2}\theta }{1 +sin^{2}\theta +cos^{2}\theta +cos^{2}\theta sin^{2}\theta}\)
= \(\frac{1-2sin^{2}\theta cos^{2}\theta+ sin^{2}\theta cos^{2}\theta\left ( cos^{2}\theta +sin^{2}\theta \right ) }{1 +1+cos^{2}\theta sin^{2}\theta}\)
= \(\frac{1-sin^{2}\theta cos^{2}\theta }{2+sin^{2}\theta cos^{2}\theta }\)
\(\therefore\)
Â
Â
Q 65 . (i) . \(\left [ \frac{1+sin\theta -cos\theta }{1+sin\theta +cos\theta} \right ]^{2}\)
Ans:
= \(\left ( \frac{1+sin\theta -cos\theta }{1+sin\theta +cos\theta}\times \frac{1+sin\theta -cos\theta}{1+sin\theta -cos\theta} \right )^{2}\)
= \(\left [ \frac{\left (1+sin\theta -cos\theta \right )^{2}}{\left ( 1+sin\theta \right )^{2}-cos^{2}\theta } \right ]^{2}\)
= \(\left [ \frac{\left ( 1 \right )^{2}+sin^{2}\theta +cos^{2}\theta +2\times 1\times sin\theta +2\times sin\theta \left ( -cos\theta \right )-2cos\theta }{1-cos^{2}\theta+sin^{2}\theta +2 sin\theta} \right ]^{2}\)
= \(\left [ \frac{ 1 +1 +2 sin\theta -2 sin\theta cos\theta -2cos\theta }{sin^{2}\theta+sin^{2}\theta +2 sin\theta} \right ]^{2}\)
= \(\left [ \frac{ 2 +2 sin\theta -2 sin\theta cos\theta -2cos\theta }{2sin^{2}\theta +2 sin\theta} \right ]^{2}\)
= \(\left [ \frac{ 2 \left (1+ sin\theta \right )-2cos\theta \left (sin\theta +1 \right ) }{2sin\theta \left (sin\theta+1 \right )} \right ]^{2}\)
= \(\left [ \frac{ \left (1+ sin\theta \right ) \left (2-2cos\theta \right ) }{2sin\theta \left (sin\theta+1 \right )} \right ]^{2}\)
= \(\left [ \frac{ \left (2-2cos\theta \right ) }{2sin\theta } \right ]^{2}\)
= \(\left [ \frac{ \left (1-cos\theta \right ) }{sin\theta } \right ]^{2}\)
= \(\frac{ \left (1-cos\theta \right )^{2} }{1-cos^{2}\theta }\)
= \(\frac{ \left (1-cos\theta \right )\times \left ( 1-cos\theta \right ) }{\left (1-cos\theta \right ) \left ( 1+cos\theta \right )}\)
= \(\frac{1-cos\theta }{1+cos\theta}\)
\(\therefore\)
Â
Â
Q 65 (ii) . \(\frac{1+sec\theta -tan\theta }{1+sec\theta +tan\theta}\)
Ans:
= LHS =Â \(\frac{1+sec\theta -tan\theta }{1+sec\theta +tan\theta}\)
= \(\frac{\left ( sec^{2 } \theta -tan^{2}\theta\right )+\left ( sec\theta -tan\theta \right )}{1+sec\theta +tan\theta }\)
= \(\frac{\left ( sec \theta -tan\theta\right )\left ( sec\theta +tan\theta \right )+\left ( sec\theta -tan\theta \right )}{1+sec\theta +tan\theta }\)
= \(\frac{\left ( sec \theta -tan\theta\right )\left (1+sec\theta +tan\theta \right )}{1+sec\theta +tan\theta }\)
= \(\left ( sec \theta -tan\theta\right )\)
= \(\frac{1}{cos\theta} -\frac{sin\theta}{cos\theta}\)
= \(\frac{1-sin\theta }{cos\theta }\)
\(\therefore\)
Â
Â
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
\(\therefore\)
Â
Â
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
\(\therefore\)
Â
Â
Q 68 . \(\left ( cosec\theta -sec\theta \right )\left ( cot\theta -tan\theta \right )\)
Ans:
LHS =Â \(\left ( cosec\theta -sec\theta \right )\left ( cot\theta -tan\theta \right )\)
\(\left [ \frac{1}{sin\theta }- \frac{1}{cos\theta }\right ]\left [ \frac{cos\theta }{sin\theta }-\frac{sin\theta }{cos\theta } \right ]\)
\(\left [ \frac{cos\theta -sin\theta }{sin\theta cos\theta }\right ]\left [ \frac{cos^{2}\theta -sin^{2}\theta}{sin\theta cos\theta } \right ]\)
\(\left [ \frac{\left (cos\theta -sin\theta \right )^{2}\left ( cos\theta +sin\theta \right )}{sin^{2}\theta cos^{2}\theta }\right ]\)
RHSÂ = \(\left ( cosec\theta +sec\theta \right )\left ( sec\theta cosec\theta – 2 \right )\)
\(\left [ \frac{1}{sin\theta }+ \frac{1}{cos\theta }\right ]\left [ \frac{1 }{cos\theta }-\frac{1 }{sin\theta }-2 \right ]\)
= \(\left [ \frac{sin\theta +cos\theta }{sin\theta cos\theta} \right ]\left [ \frac{1-2 sin\theta cos\theta}{sin\theta cos\theta} \right ]\)
= \(\left [ \frac{sin\theta +cos\theta }{sin\theta cos\theta} \right ]\left [ \frac{cos^{2}\theta +sin^{2}\theta -2 sin\theta cos\theta}{sin\theta cos\theta} \right ]\)
= \(\left [ \frac{\left (cos\theta -sin\theta \right )^{2}\left ( cos\theta +sin\theta \right )}{sin^{2}\theta cos^{2}\theta }\right ]\)
\(\therefore\)
Q 70 . \(\frac{cos A cosec A-sin A sec A}{cos A +sin A}\)
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
\(\therefore\)
Â
Â
Q 71 . \(\frac{sin A}{sec A +tan A -1}+\frac{cos A}{cosec A + cot A-1}\)
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
\(\therefore\)
Â
Â
Q 72 . \(\frac{tan A}{\left (1 +tan^{2}A \right )^{2}}+\frac{cot A}{\left (1 +cot^{2}A \right )^{2}}\)
Ans:
\(\frac{tan A}{\left (sec^{2}A \right )^{2}}+\frac{cot A}{\left (cosec^{2}A \right )^{2}}\)
= \(\frac{\frac{sin A}{cos A}}{sec^{4}A }+\frac{\frac{cos A}{sin A}}{cosec^{4}A }\)
= \(\frac{\frac{sin A}{cos A}}{\frac{1}{cos^{4}A} }+\frac{\frac{cos A}{sin A}}{\frac{1}{sin^{4}A} }\)
= \(\frac{sin A}{cosA}\times \frac{cos^{4}A}{1}+\frac{cos A }{sin A}\times \frac{sin ^{4}A}{1}\)
= \(sin A\times cos^{3}A+cos A \times sin ^{3}A\)
= \(sin A cosA\left ( cos^{2}A+sin ^{2}A \right )\)
= \(sin A cosA\)
\(\therefore\)
Â
Â
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
= \(sec^{4}A\;-\;sec^{4}A\;\times \;sin^{4}A\;-\;2tan^{4}A\)
= \(sec^{4}A\;-\;(\frac{1}{cos^{4}A}\;\times \;sin^{4}A)\;-\;2tan^{4}A\)
= \(sec^{4}A\;-\;tan^{4}A\;-\;2tan^{4}A\)
= \((sec^{2}A)^{2}\;-\;tan^{4}A\;-\;2tan^{4}A\)
= \((1\;+\;tan^{2}A)^{2}\;-\;tan^{4}A\;-\;2tan^{4}A\)
= \(1+tan^{4}A+2tan^{2}A-\;tan^{4}A\;-\;2tan^{4}A\)
= 1
Hence, L.H.S = R.H.S
Â
Â
Q74. \(\frac{cot^{2}A(secA\;-\;1)}{1\;+\;sinA}\)
Ans:
Given, L.H.S = \(\frac{cot^{2}A(secA\;-\;1)}{1\;+\;sinA}\)
Here, \(sin^{2}A\;+\;cos^{2}A\)
= \(\frac{\frac{cos^{2}A}{sin^{2}A}(\frac{1}{cosA}-1)}{1+sinA}\)
= \(\frac{\frac{cos^{2}A}{sin^{2}A}(\frac{1-cosA}{cosA})}{1+sinA}\)
= \(\frac{\frac{cosA\times cosA}{(1-cos^{2}A)}(\frac{1-cosA}{cosA})}{1+sinA}\)
= \(\frac{(cosA)}{(1+cosA)}\frac{1}{1+sinA}\)
Solving,
RHS =>Â \(sec^{2}a[\frac{1-sinA}{1+secA}]\)
= \(\frac{1}{cos^{2}A}[\frac{1-sinA}{1+secA}]\)
= \(\frac{1}{cos^{2}A}[\frac{1-sinA}{1+secA}]\)
= \(\frac{1}{cos^{2}A}[\frac{1-sinA}{cosA+1}]cosA\)
= \(\frac{(1-sinA)}{(cosA+1)(cosA)}\)
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
Â
Â
Q75. \((1\;+\;cotA\;+tanA)(sinA\;-\;cosA)\)
Ans:
Given, L.H.S = \((1\;+\;cotA\;+tanA)(sinA\;-\;cosA)\)
=> sinA – cosA + cotAsinA – cotAcosA + sinAtanA – tanAcosA
=> sinA – cosA + \(\frac{cosA}{sinA}\times sinA\)
=> sinA – cosA + cosA – cotAcosA + sinAtanA – sinA
=> sinAtanA – cosAcotA
=> \(\frac{secA}{cosec^{2}A}\)
Here, secA = \(\frac{1}{cosA}\)
=> \(\frac{sin^{2}A}{cosA}\)
=> \(\frac{sin^{2}A\;-\;cos^{2}A}{cosAsinA}\)
=> \((sinA\times \frac{sinA}{cosA})\)
=> sinAtanA – cosAcotA
Hence, L.H.S = R.H.S
Â
Â
Q76. If \(\frac{x}{a}cos\theta \;+\;\frac{y}{b}sin\theta\)
Ans:
Given,
=> (\(\frac{x}{a}cos\theta \;+\;\frac{y}{b}sin\theta\)
=> \(\frac{x^{2}}{a^{2}}cos^{2}\theta \;+\;\frac{y^{2}}{b^{2}}sin^{2}\theta \;+\;\frac{2xy}{ab}cos\theta sin\theta \;+\;\frac{x^{2}}{a^{2}}sin^{2}\theta \;+\;\frac{y^{2}}{b^{2}}\;-\;\frac{2xy}{ab}sin\theta cos\theta\)
=> \(\frac{x^{2}}{a^{2}}cos^{2}\theta \;+\;\frac{y^{2}}{b^{2}}sin^{2}\theta \;+\;\frac{x^{2}}{a^{2}}sin^{2}\theta \;+\;\frac{y^{2}}{b^{2}}sin^{2}\theta\)
=> \(cos^{2}\theta [\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}]\)
=> \((cos^{2}\theta\;+\;sin^{2}\theta ) [\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}]\)
Here cos2A +sin2A = 1
=> (1) [\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}]\)
Â
Â
Q77. If \(cosec\theta \;-\;sin\theta \;=\;a^{3}\)
Ans:
Given, \(cosec\theta \;-\;sin\theta \;=\;a^{3}\)
Here, \(cosec\theta \;=\;\frac{1}{sin\theta }\)
=> \(\frac{1}{sin\theta }\)
=> \(\frac{1\;-\;sin^{2}\theta }{sin\theta }\)
Here cos2A +sin2A = 1
=> \(\frac{cos^{2}\theta }{sin\theta }\)
=> \(\frac{cos^{\frac{2}{3}}\theta }{sin^{\frac{1}{3}}\theta }\)
Squaring on both sides
=> a2 = \(\frac{cos^{\frac{4}{3}}\theta }{sin^{\frac{2}{3}}\theta }\)
\(sec\theta \;-\;cos\theta \;=\;b^{3}\)
=> \(\frac{1}{cos\theta }\)
=> \(\frac{1\;-\;cos^{2}\theta }{cos\theta }\)
=> \(\frac{sin^{2}\theta }{cos\theta }\)
=> \(\frac{sin^{\frac{2}{3}}\theta }{cos^{\frac{1}{3}}\theta }\)
Squaring on both sides
=> b2 = \(\frac{sin^{\frac{4}{3}}\theta }{cos^{\frac{2}{3}}\theta }\)
Now, \(a^{2}b^{2}(a^{2}\;+\;b^{2})\)
=> \(\frac{cos^{\frac{4}{3}}\theta }{sin^{\frac{2}{3}}\theta }\)
=> \(cos^{\frac{2}{3}}\theta\)
= 1
Hence, L.H.S = R.H.S
Â
Â
Q78. If acos3 \(\theta\)
Ans:
Given, \((m\;+\;n)^{\frac{2}{3}}\)
Substitute the values of m and n in the above equation
=> (\(( acos3 \(\theta\)
=> \((a)^{\frac{2}{3}}\)
=> \((a)^{\frac{2}{3}}\)
=> \((a)^{\frac{2}{3}}\)
=> \((a)^{\frac{2}{3}}\)
=>Â \((a)^{\frac{2}{3}}\)
=> \((a)^{\frac{2}{3}}\)
=> \((a)^{\frac{2}{3}}\)
=> 2\((a)^{\frac{2}{3}}\)
Hence, L.H.S = R.H.S
Â
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.
Â
Â
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
Â
Â
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.
Â
Â
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:
\(cos\Theta+cos^{2}\Theta=1\)
\(cos\Theta=1-cos^{2}\Theta\)
\(cos\Theta=sin^{2}\Theta\)
\(Now,\;sin^{12}\Theta+3sin^{10}\Theta+3sin^{8}\Theta+sin^{6}\Theta+2sin^{4}\Theta+2sin^{2}\Theta-2\)
\(=(sin^{4}\Theta)^{3}+3sin^{4}\Theta.sin^{2}\Theta(sin^{4}\Theta+sin^{2}\Theta)+(sin^{2}\Theta)^{3}+2(sin^{2}\Theta)^{2}+2sin^{2}\Theta-2\)
\(Using\;(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)\;and\;also\;from\;(i)\;cos\Theta=sin^{2}\Theta\)
\((sin^{4}\Theta+sin^{2}\Theta)^{3}+2cos^{2}\Theta+2cos\Theta-2\)
\(((sin^{2}\Theta)^{2}+sin^{2}\Theta)^{3}+2cos^{2}\Theta+2cos\Theta-2\)
\((cos^{2}\Theta+sin^{2}\Theta)^{3}+2cos^{2}\Theta+2cos\Theta-2\)
\(1+2cos^{2}\Theta+2sin^{2}\Theta-2\;\;\;\;\;[\because sin^{2}\Theta+cos^{2}\Theta=1]\)
\(1+2(cos^{2}\Theta+sin^{2}\Theta)-2\)
\(1+2(1)-2\)
\(=1\)
L.H.S = R.H.S
Hence proved.
Q83: Given that: \((1+cos\alpha)(1+cos\beta)(1+cos\gamma)=(1-cos\alpha)(1-cos\beta)(1-cos\gamma)\)
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 \(sin\alpha\;sin\beta\;sin\gamma\)
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
\((sin\Theta+cos\Theta)^{2}=x^{2}\)
\(\Rightarrow sin\Theta^{2}+cos\Theta^{2}+2sin\Theta cos\Theta=x^{2}\)
\(\therefore sin\Theta cos\Theta=\frac{x^{2}-1}{2}……(i)\)
\(We\;know\;that\;sin^{2}\Theta+cos^{2}\Theta=1\)
Cubing on both sides
\((sin^{2}\Theta+cos^{2}\Theta)^{3}=1^{3}\)
\(sin^{6}\Theta+cos^{6}\Theta+3sin^{2}\Theta cos^{2}\Theta(sin^{2}\Theta+cos^{2}\Theta)=1\)
\(\Rightarrow sin^{6}\Theta+cos^{6}\Theta=1-3sin^{2}\Theta cos^{2}\Theta\)
\(\Rightarrow sin^{6}\Theta+cos^{6}\Theta=1-\frac{3(x^{2}-1)^{2}}{4}\)
\(\therefore sin^{6}\Theta+cos^{6}\Theta=\frac{4-3(x^{2}-1)^{2}}{4}\)
Â
Q85. If x = \(asec\theta cos\phi\)
Ans:
Given, x = \(asec\theta cos\phi\)
y =\(bsec\theta sin\phi\)
z= \(ctan\phi\)
squaring x,y,z on the sides
\(x^{2}\)
\(\frac{x^{2}}{a^{2}}\)
\(y^{2}\)
\(\frac{y^{2}}{b^{2}}\)
\(z^{2}\)
\(\frac{z^{2}}{c^{2}}\)
Substitute eq 1,2,3 in  \(\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}\;-\;\frac{z^{2}}{c^{2}}\)
=> \(\frac{x^{2}}{a^{2}}\;+\;\frac{y^{2}}{b^{2}}\;-\;\frac{z^{2}}{c^{2}}\)
=> \(sec^{2}\theta cos^{2}\phi\)
=> \(sec^{2}\theta (cos^{2}\phi \;+\;sin^{2}\phi )\)
We know that, \(cos^{2}\phi \;+\;sin^{2}\phi\)
=> \(sec^{2}\theta\)
And, \(sec^{2}\theta \;-\;tan^{2}\theta\)
=> 1
Hence, L.H.S= R.H.S
Â
Â
Q86. If \(sin\theta \;+\;2cos\theta\)
Ans:
Given, \(sin\theta \;+\;2cos\theta\)
Squaring on both sides
=> \((sin\theta \;+\;2cos\theta)^{2}\)
=> \(sin^{2}\theta \;+\;4cos^{2}\theta\;+\;4sin\theta cos\theta\)
=>Â \(4cos^{2}\theta\;+\;4sin\theta cos\theta\)
Here, 1 – \(sin^{2}\theta\)
=> \(4cos^{2}\theta\;+\;4sin\theta cos\theta\)
=> \(3cos^{2}\theta\;+\;4sin\theta cos\theta\)
We have, \(2sin\theta \;-\;cos\theta\)
Squaring L.H.S
\((2sin\theta\;-\;cos\theta )^{2}\)
Here, \(4sin\theta cos\theta\)
= \(4sin^{2}\theta \;+\;cos^{2}\theta \;+\;3cos^{2}\theta\)
= \(4sin^{2}\theta \;+\;4cos^{2}\theta\)
= \(4(sin^{2}\theta \;+\;cos^{2}\theta)\)
=Â 4(1)
= 4
\((2sin\theta\;-\;cos\theta )^{2}\)
=> \(2sin\theta \;-\;cos\theta\)
Hence proved