Question

square root of 1 + cos theta by 1 - cos theta is equal to cosec theta + cot theta

Answer 1

Solution:

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

Multiply numerator and denominator by $(1+cos\theta)$, 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:

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

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

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

= $cosec\theta+cot\theta$

=$RHS$

Therefore,

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

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Answer 2

Answer:

I hope it wil help you.....☺