Question

Answer this!

Required QUESTION..​

Answer 1

LHS = RHS

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Answer 2

Solution:-

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

Multiply numerator and denominator by (1+cosθ)(1+cosθ) , we get

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

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

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

We know the Trigonometric identity:-

$\boxed {1-cos^{2}\theta = sin^{2}\theta}$

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

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

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