Question

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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

Answer 2

Answer:

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