Question

Prove the following trignometric identities:

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

Answer 1

We know ,

$co{s}^2\theta +si{n}^2\theta =1$

Multiplying both numerator and denominator by (1−cosθ), we have

$\sqrt{\frac{1-cos\theta }{1+cos\theta }}$
= $\sqrt{\frac{(1-cos\theta )(1-cos\theta )}{(1+cos\theta )(1-cos\theta )}}$
= $\sqrt{\frac{(1-cos\theta {)}^2}{(1-co{s}^2\theta )}}$
= $\sqrt{\frac{(1-cos\theta {)}^2}{\left(si{n}^2\theta \right)}}$
= $\frac{(1-cos\theta )}{sin\theta }$
= $\frac{1}{sin\theta }$ - $\frac{cos\theta }{sin\theta }$

= $cosec\theta -cot\theta$