RD Sharma Solutions Class 10 Trigonometric Identities Exercise 6.1

RD Sharma Solutions Class 10 Chapter 6 Exercise 6.1

RD Sharma Class 10 Solutions Chapter 6 Ex 6.1 PDF Free Download

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,

$tan\theta+ \frac{1}{tan\theta} =\frac{tan^{2}\theta+1}{tan\theta}$

$=\frac{sec^{2}\theta}{tan\theta}$

$=sec\theta\frac{sec\theta}{tan\theta}$

$=sec\theta\frac{\frac{1}{cos\theta}}{\frac{sin\theta}{cos\theta}}$

$=sec\theta\frac{1}{sin\theta}$

$=sec\thetacosec\theta$

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: $$\frac{1-cos\theta }{sin\theta}=\frac{sin\theta}{1+cos\theta}$$\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}$$sin^{2}\theta$ => $sin^{2}\theta$[ $\frac{1}{cos^{2}\theta}$ – 1] => $sin^{2}\theta$[ $\frac{1\;-\;cos^{2}\theta}{cos^{2}\theta}$] => $\frac{sin^{2}\theta}{cos^{2}\theta}$ * $sin^{2}\theta$ => $tan^{2}\theta\;*\;sin^{2}\theta$ Hence, L.H.S = R.H.S Q17. (cosec $\theta$ + sin $\theta$)(cosec $\theta$ – sin $\theta$) = $cot^{2}\theta\;+\;cos^{2}\theta$ Ans: Given, L.H.S = (cosec $\theta$ + sin $\theta$)(cosec $\theta$ – sin $\theta$) Here, (a + b)(a – b) = a2 – b2 cosec2 $\theta$ can be written as 1 + cot2 $\theta$ and sin2 $\theta$ can be written as 1 – cos2 $\theta$ => 1 + cot2 $\theta$ – (1 – cos2 $\theta$) => 1 + cot2 $\theta$ – 1 + cos2 $\theta$ => cot2 $\theta$ + cos2 $\theta$ Hence, L.H.S = R.H.S Q18. ($sec\theta$ +cos $\theta$)(sec $\theta$ – cos $\theta$) = $tan^{2}\theta\;+\;sin^{2}\theta$ Ans: Given, L.H.S = (sec $\theta$ + cos $\theta$)(sec $\theta$ – cos $\theta$) Here, (a + b)(a – b) = a2 – b2 sec2$\theta$ can be written as 1 + tan2 $\theta$ and cos2 $\theta$ can be written as 1 – sin2 $\theta$ => 1 + tan2 $\theta$ – (1 – sin2 $\theta$) => 1 + tan2 $\theta$ – 1 + sin2 $\theta$ => tan2 $\theta$ + sin2 $\theta$ Hence, L.H.S = R.H.S Q19. secA(1 – sinA)(secA + tanA) = 1 Ans: Given, L.H.S = secA(1 – sinA)(secA + tanA) Here, secA = $\frac{1}{cosA}$ and tanA = $\frac{sinA}{cosA}$ => $\frac{1}{cosA}$ * (1 – sinA) * $\frac{1 + sinA}{cosA}$ => $\frac{cos2A}{cos2A}$$\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)$$(sin^{2}\A\;*\;cot^{2}A)\;+\;(cos^{2}A\;*\;tan^{2}A)\]”> = 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$$\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] $\Rightarrow 2sec^{2}A$ $\therefore$ 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 $sec\theta$ $\therefore$ 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 = $\frac{\left ( 1+sin\theta \right )^{2}+\left ( 1-sin\theta \right )^{2}}{2cos^{2}\theta }\\=\frac{\left ( 1+2sin\theta+sin^{2} \theta \right )+\left ( 1-2sin\theta+sin^{2} \theta \right )}{2cos^{2}\theta}\\=\frac{ 1+2sin\theta+sin^{2} \theta +1-2sin\theta+sin^{2} \theta }{2cos^{2}\theta}\\=\frac{2+2sin^{2}\theta }{2cos^{2}\theta }\\=\frac{2\left ( 1+sin^{2}\theta \right )}{2\left ( 1-sin^{2}\theta \right )}\\=\frac{\left ( 1+sin^{2}\theta \right )}{\left ( 1-sin^{2}\theta \right )}$ $\therefore$ 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 + $cot^{2}\theta$ = $cosec^{2}\theta$] = $\frac{1}{cos^{2}\theta \cdot 1}sin^{2}\theta$ = $tan^{2}\theta$ $\therefore$ 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 = $\frac{sin^{2}\theta }{1-cos\theta }$ = $\frac{1 – cos^{2}\theta }{1-cos\theta }$ = $\frac{\left ( 1-cos\theta \right )\left ( 1+cos\theta \right )}{1-cos\theta }$ = $1 + cos\theta$ $\therefore$ 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 , $a^{3}-b^{3}=\left ( a-b \right )\left ( a^{2}+ab+b^{2} \right )$] = $\frac{ 1+tan\theta +tan^{2}\theta }{tan\theta}$ = $\frac{1}{tan\theta }+\frac{tan\theta}{tan\theta}+\frac{tan^{2}\theta}{tan\theta}$ $1 + tan\theta +cot\theta$ $\therefore$ 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 = $\frac{sec^{2}\theta \cdot cot\theta }{cosec^{2}\theta }$ = $\frac{1\cdot sin^{2}\theta }{cos^{2}\theta }\cdot \frac{cos\theta }{sin\theta }$ $\left [ \because sec\theta = \frac{1}{cos\theta } , cosec\theta = \frac{1}{sin\theta }, cot \theta = \frac{cos \theta }{sin\theta }\right ]$ $\Rightarrow \frac{sin\theta }{cos\theta }= tan\theta$ $\therefore$ 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 $sin^{2}A=1-cos^{2}A$ $\Rightarrow sin^{2}A=\left (1-cosA \right )\left ( 1+cosA \right )$ $\Rightarrow\;LHS = \frac{\left ( 1+cosA \right )}{\left ( 1-cosA \right )\left ( 1+cosA \right )}$ = $\Rightarrow\;LHS = \frac{1}{\left ( 1-cosA \right )}$ $\therefore$ 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 $\frac{sec A-tanA}{sec A+tanA}\times \frac{sec A+tanA}{sec A+tanA}$ = $\frac{sec^{2} A-tan^{2}A}{\left (sec A+tanA \right )^{2}}$ = $\frac{1}{\left (sec A+tanA \right )^{2}}$ = $\frac{1}{\left (sec ^{2}A+tan^{2}A +2sec A tan A \right )}$ = $\frac{1}{\left (\frac{1}{cos ^{2}A}+\frac{sin ^{2}A}{cos ^{2}A} +\frac{2sinA}{cosA} \right )}$ $\Rightarrow \frac{cos ^{2}A}{1 +sin^{2}A+2 sin A}$ = $\frac{cos ^{2}A}{\left (1 +sinA \right )^{2}}$ $\therefore$ 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 , $\frac{\left (1+cosA \right )\left ( 1-cosA \right )}{sin A\left (1-cosA \right )}$ = $\frac{1-cos^{2}A }{sin A\left (1-cosA \right )}$ = $\frac{sin^{2}A }{sin A\left (1-cosA \right )}$ = $\frac{sinA }{1-cosA }$ $\therefore$ 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 1. 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 1. (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 1. $\frac{1 – cos A}{1 + cos A}$ = (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 1. $\frac{1}{sec A – 1}+\frac{1}{sec A+1} = 2 cosec A cot A$ 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 1. $\frac{cos A}{1-tan A}+\frac{sin A}{1-cot A}$ = 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 1. $\frac{(cosec A)}{(cosec A-1)}+\frac{(cosec A)}{(cosec A+1)}$ = 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 1. $\frac{cot A-cos A}{cot A+cos A}$ = $\frac{cosec A-1}{cosec A + 1}$ 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 1. $\frac{1+cos\Theta -sin^{2}\Theta }{sin\Theta (1+cos\Theta)}$ = 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, = $\frac{\frac{1+cos\Theta +sin\Theta}{cos\Theta }}{\frac{1+cos\Theta -sin\Theta}{cos\Theta }}$ = $\frac{sec\Theta+1+tan\Theta }{sec\Theta+1-tan\Theta }$ = $\frac{1+sec\Theta+tan\Theta }{1+sec\Theta-tan\Theta }$ [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) $\frac{sin\Theta-cos\Theta+1 }{sin\Theta+cos\Theta-1}$ = $\frac{1}{sec\Theta-tan\Theta}$ 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, = $\frac{tan\Theta+sec\Theta-1 }{tan\Theta-sec\Theta+1}$ [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 (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, = $\frac{\frac{cos\Theta-sin\Theta+1}{sin\Theta } }{\frac{cos\Theta+sin\Theta-1}{sin\Theta}}$ = $\frac{cot\Theta+cosec\Theta-1}{cot\Theta-cosec\Theta+1}$ [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) $(sin\Theta + cos\Theta )(tan\Theta + cot\Theta)$ = $sec\Theta+cosec\Theta$ Ans: To prove, $(sin\Theta + cos\Theta )(tan\Theta + cot\Theta)$ = $cosec\Theta+cosec\Theta$ Considering LHS, we get, = $(sin\Theta + cos\Theta)(\frac{sin\Theta}{cos\Theta }+\frac{cos\Theta}{sin\Theta})$ = $(\frac{sin^{2}\Theta}{cos\Theta}+ cos\Theta+sin\Theta+\frac{cos^{2}\Theta}{sin\Theta}$ = $sin\Theta(tan\Theta +1) +cos\Theta( \frac{1}{tan\Theta}+1)$ = $sin\Theta(tan\Theta +1) +\frac{cos\Theta}{tan\Theta}(tan\Theta +1)$ = $(sin\Theta+\frac{cos\Theta}{tan\Theta})(tan\Theta +1)$ = $(\frac{sin^{2}\Theta+cos^{2}\Theta}{sin\Theta})(tan\Theta +1)$ = $(\frac{1}{sin\Theta})(tan\Theta +1)$ = $(\frac{sin\Theta +cos\Theta }{cos\Thetasin\Theta })$ = $sec\Theta+cosec\Theta$ Therefore, LHS = RHS Hence proved 1. \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 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. 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 = $\frac{\left (sin^{5}\theta +cos^{5}\theta \right ) – \left (sin^{7}\theta +cos^{7}\theta \right ) }{sin^{3}\theta +cos^{3]\theta }$ = $\frac{sin^{5}\theta \left ( 1- sin^{2}\theta \right )+cos^{5}\theta \left ( 1- cos^{2}\theta \right ) }{sin^{3]\theta +cos^{3}\theta }$ = $\frac{sin^{5}\theta \times cos^{2}\theta +cos^{5}\theta \times sin^{2}\theta }{sin^{3}\theta +cos^{3}\theta }$ = $\frac{sin^{2}\theta cos^{2}\theta \left ( sin^{3}\theta +cos^{3}\theta \right ) }{sin\theta +cos\theta }$ = $sin^{2}\theta cos^{2}\theta$ $\therefore$ 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 $\therefore$ 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: $\left [ \frac{1}{\frac{1}{cos^{2}\theta }-cos^{2}\theta } +\frac{1}{\frac{1}{sin^{2}\theta }-sin^{2}\theta}\right ]sin^{2}\theta cos^{2}$ = $\left [ \frac{1}{\frac{1-cos^{4}\theta}{cos^{2}\theta } } +\frac{1}{\frac{1-sin^{4}\theta }{sin^{2}\theta}}\right ]sin^{2}\theta cos^{2}\theta$ = $\left [ \frac{cos^{2}\theta }{1-cos^{4}\theta} +\frac{sin^{2}\theta}{1-sin^{4}\theta}\right ]sin^{2}\theta cos^{2}\theta$ = $\left [ \frac{cos^{2}\theta }{cos^{2}\theta+sin^{2}\theta-cos^{4}\theta} +\frac{sin^{2}\theta}{cos^{2}\theta+sin^{2}\theta-sin^{4}\theta}\right ]sin^{2}\theta cos^{2}\theta$ = $\left [ \frac{cos^{2}\theta }{cos^{2}\theta\left ( 1-cos^{2}\theta \right )+sin^{2}\theta} +\frac{sin^{2}\theta}{cos^{2}\theta+sin^{2}\theta\left ( 1- sin^{2}\theta\right )}\right ]sin^{2}\theta cos^{2}\theta$ = $\left [ \frac{cos^{2}\theta }{cos^{2}\theta sin^{2}\theta+sin^{2}\theta} +\frac{sin^{2}\theta}{cos^{2}\theta+sin^{2}\theta cos^{2}\theta}\right ]sin^{2}\theta cos^{2}\theta$ = $\left [ \frac{cos^{2}\theta }{sin^{2}\theta \left ( cos^{2}\theta +1 \right )} +\frac{sin^{2}\theta}{cos^{2}\theta\left ( sin^{2}\theta+1 \right )}\right ]sin^{2}\theta cos^{2}\theta$ = $\frac{cos^{4}\theta \left ( sin^{2}\theta+1 \right )+sin^{4}\theta\left ( cos^{2}\theta +1 \right )}{sin^{2}\theta cos^{2}\theta \left ( cos^{2}\theta +1 \right )\left (sin^{2}\theta+1 \right ) } sin^{2}\theta cos^{2}\theta$ = $\frac{cos^{4}\theta \left ( sin^{2}\theta+1 \right )+sin^{4}\theta\left ( cos^{2}\theta +1 \right )}{ \left ( cos^{2}\theta +1 \right )\left (sin^{2}\theta+1 \right ) }$ = $\frac{cos^{4}\theta +cos^{4}\theta sin^{2}\theta +sin^{4}\theta +sin^{4}\theta cos^{2}\theta }{1 +sin^{2}\theta +cos^{2}\theta +cos^{2}\theta sin^{2}\theta}$ = $\frac{1-2sin^{2}\theta cos^{2}\theta+ sin^{2}\theta cos^{2}\theta\left ( cos^{2}\theta +sin^{2}\theta \right ) }{1 +1+cos^{2}\theta sin^{2}\theta}$ = $\frac{1-sin^{2}\theta cos^{2}\theta }{2+sin^{2}\theta cos^{2}\theta }$ $\therefore$ 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: = $\left ( \frac{1+sin\theta -cos\theta }{1+sin\theta +cos\theta}\times \frac{1+sin\theta -cos\theta}{1+sin\theta -cos\theta} \right )^{2}$ = $\left [ \frac{\left (1+sin\theta -cos\theta \right )^{2}}{\left ( 1+sin\theta \right )^{2}-cos^{2}\theta } \right ]^{2}$ = $\left [ \frac{\left ( 1 \right )^{2}+sin^{2}\theta +cos^{2}\theta +2\times 1\times sin\theta +2\times sin\theta \left ( -cos\theta \right )-2cos\theta }{1-cos^{2}\theta+sin^{2}\theta +2 sin\theta} \right ]^{2}$ = $\left [ \frac{ 1 +1 +2 sin\theta -2 sin\theta cos\theta -2cos\theta }{sin^{2}\theta+sin^{2}\theta +2 sin\theta} \right ]^{2}$ = $\left [ \frac{ 2 +2 sin\theta -2 sin\theta cos\theta -2cos\theta }{2sin^{2}\theta +2 sin\theta} \right ]^{2}$ = $\left [ \frac{ 2 \left (1+ sin\theta \right )-2cos\theta \left (sin\theta +1 \right ) }{2sin\theta \left (sin\theta+1 \right )} \right ]^{2}$ = $\left [ \frac{ \left (1+ sin\theta \right ) \left (2-2cos\theta \right ) }{2sin\theta \left (sin\theta+1 \right )} \right ]^{2}$ = $\left [ \frac{ \left (2-2cos\theta \right ) }{2sin\theta } \right ]^{2}$ = $\left [ \frac{ \left (1-cos\theta \right ) }{sin\theta } \right ]^{2}$ = $\frac{ \left (1-cos\theta \right )^{2} }{1-cos^{2}\theta }$ = $\frac{ \left (1-cos\theta \right )\times \left ( 1-cos\theta \right ) }{\left (1-cos\theta \right ) \left ( 1+cos\theta \right )}$ = $\frac{1-cos\theta }{1+cos\theta}$ $\therefore$ 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 , $sec^{2}\theta -tan^{2}\theta = 1$] = $\frac{\left ( sec \theta -tan\theta\right )\left ( sec\theta +tan\theta \right )+\left ( sec\theta -tan\theta \right )}{1+sec\theta +tan\theta }$ = $\frac{\left ( sec \theta -tan\theta\right )\left (1+sec\theta +tan\theta \right )}{1+sec\theta +tan\theta }$ = $\left ( sec \theta -tan\theta\right )$ = $\frac{1}{cos\theta} -\frac{sin\theta}{cos\theta}$ = $\frac{1-sin\theta }{cos\theta }$ $\therefore$ 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 $\therefore$ 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 $\therefore$ 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 = $\left ( cosec\theta +sec\theta \right )\left ( sec\theta cosec\theta – 2 \right )$ $\left [ \frac{1}{sin\theta }+ \frac{1}{cos\theta }\right ]\left [ \frac{1 }{cos\theta }-\frac{1 }{sin\theta }-2 \right ]$ = $\left [ \frac{sin\theta +cos\theta }{sin\theta cos\theta} \right ]\left [ \frac{1-2 sin\theta cos\theta}{sin\theta cos\theta} \right ]$ = $\left [ \frac{sin\theta +cos\theta }{sin\theta cos\theta} \right ]\left [ \frac{cos^{2}\theta +sin^{2}\theta -2 sin\theta cos\theta}{sin\theta cos\theta} \right ]$ = $\left [ \frac{\left (cos\theta -sin\theta \right )^{2}\left ( cos\theta +sin\theta \right )}{sin^{2}\theta cos^{2}\theta }\right ]$ $\left [ \because cos^{2}\theta +sin^{2}\theta = 1 \right ]$ $\therefore$ 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 $\therefore$ 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 $\therefore$ 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 ] = $\frac{\frac{sin A}{cos A}}{sec^{4}A }+\frac{\frac{cos A}{sin A}}{cosec^{4}A }$ = $\frac{\frac{sin A}{cos A}}{\frac{1}{cos^{4}A} }+\frac{\frac{cos A}{sin A}}{\frac{1}{sin^{4}A} }$ = $\frac{sin A}{cosA}\times \frac{cos^{4}A}{1}+\frac{cos A }{sin A}\times \frac{sin ^{4}A}{1}$ = $sin A\times cos^{3}A+cos A \times sin ^{3}A$ = $sin A cosA\left ( cos^{2}A+sin ^{2}A \right )$ = $sin A cosA$ $\therefore$ 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

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

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

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

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

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

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

Cubing on both sides

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

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

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

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

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

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

Practise This Question

An n-digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is