Question

cosθ/1-sinθ + sinθ/1-cosθ = secθ + cosecθ + secθ*cosecθ
plsss answer it as fast as u can
correct will be marked brainliest

Answer 1

Answer:

$LHS = \frac{cos\theta}{1-sin\theta}+ \frac{sin\theta}{1-cos\theta}$

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

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

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

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

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

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

$= sec\theta + cosec\theta + \frac{sin^{2}\theta+cos^{2}\theta}{sin\theta cos\theta}$

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

$= sec\theta + cosec\theta + \frac{1}{sin\theta}\cdot \frac{1}{cos\theta}$

$= sec\theta + cosec\theta + cosec\theta sec\theta$

$= RHS$

$Hence,Proved$

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Answer 2

Answer:

LHS = \frac{cos\theta}{1-sin\theta}+ \frac{sin\theta}{1-cos\theta}LHS=1−sinθcosθ+1−cosθsinθ

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

= \frac{cos\theta(1+sin\theta)}{1^{2}-sin^{2}\theta}+ \frac{sin\theta(1+cos\theta)}{1^{2}-cos^{2}\theta}=12−sin2θcosθ(1+sinθ)+12−cos2θsinθ(1+cosθ)

= \frac{cos\theta(1+sin\theta)}{cos^{2}\theta}+ \frac{sin\theta(1+cos\theta)}{sin^{2}\theta}=cos2θcosθ(1+sinθ)+sin2θsinθ(1+cosθ)

= \frac{1+sin\theta}{cos\theta}+ \frac{1+cos\theta}{sin\theta}=cosθ1+sinθ+sinθ1+cosθ

= \frac{1}{cos\theta} + \frac{sin\theta}{cos\theta} + \frac{1}{sin\theta}+ \frac{cos\theta}{sin\theta}=cosθ1+cosθsinθ+sinθ1+sinθcosθ

= \frac{1}{cos\theta} + \frac{1}{sin\theta}+\frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta}=cosθ1+sinθ1+cosθsinθ+sinθcosθ

= sec\theta + cosec\theta + \frac{sin^{2}\theta+cos^{2}\theta}{sin\theta cos\theta}=secθ+cosecθ+sinθcosθsin2θ+cos2θ

= sec\theta + cosec\theta + \frac{1}{sin\theta cos\theta}=secθ+cosecθ+sinθcosθ1

= sec\theta + cosec\theta + \frac{1}{sin\theta}\cdot \frac{1}{cos\theta}=secθ+cosecθ+sinθ1⋅cosθ1

= sec\theta + cosec\theta + cosec\theta sec\theta=secθ+cosecθ+cosecθsecθ

= RHS=RHS

Hence,ProvedHence,Proved

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