Question

Prove :- $\frac{\sec \theta +\tan \theta }{\cos ec\theta +\cot \theta }=\frac{\cos ec\theta -\cot \theta }{\sec \theta -\tan \theta }$

Answer 1

L.H.S

$\frac{\sec \theta +\tan \theta }{\cos ec\theta +\cot \theta }$

$=(\sec \theta +\tan \theta )\times \frac{(\sec \theta -\tan \theta )}{(\sec \theta -\tan \theta )}\times \frac{1}{(\cos ec\theta +\cot \theta )}\times \frac{(\cos ec\theta -\cot \theta )}{(\cos ec\theta -\cot \theta )}$

$=\frac{({\sec }^2\theta -{\tan }^2\theta )}{(\sec \theta -\tan \theta )}\times \frac{(\cos ec\theta -\cot \theta )}{(\cos e{c}^2\theta -{\cot }^2\theta )}$

We know that

${\sec }^2\theta -{\tan }^2\theta =1$

$\cos e{c}^2\theta -{\cot }^2\theta =1$

Therefore,

$=\frac{1}{(\sec \theta -\tan \theta )}\times \frac{(\cos ec\theta -\cot \theta )}{1}$

$=\frac{(\cos ec\theta -\cot \theta )}{(\sec \theta -\tan \theta )}$

R.H.S

Hence, proved.