Prove that :

Prove that :
$\sqrt{\frac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\frac{\sec \theta +1}{\sec \theta -1}}= 2\, cosec\, \theta$

Answers (1)

To Prove $\rightarrow \sqrt{\frac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\frac{\sec \theta +1}{\sec \theta -1}}= 2\, cosec\, \theta$
Proof $\rightarrow$ Taking L.H.S
$\sqrt{\frac{\left ( \sec \theta -1 \right )}{\left ( \sec \theta +1 \right )}\times \frac{\left ( \sec \theta -1 \right )}{\left ( \sec \theta -1 \right )}}+\sqrt{\frac{\left ( \sec \theta +1 \right )}{\left ( \sec \theta -1 \right )}\times \frac{\left ( \sec \theta +1 \right )}{\left ( \sec \theta +1 \right )}}$          ( $\therefore$ multiplying and dividing to make the denominater common ) $\Rightarrow \sqrt{\frac{\left ( \sec \theta -1 \right )^{2}}{\left ( \sec ^{2}\theta -1 \right )}}+\sqrt{\frac{\left ( \sec \theta +1^{2} \right )^{2}}{\left ( \sec ^{2}\theta -1^{2} \right )}}$              $\left [ \left ( a+b \right )\left ( a-b \right )= a^{2}-b^{2} \right ]$
$\Rightarrow \frac{\left ( \sec \theta -1 \right )+\left ( \sec \theta +1 \right )}{\sqrt{\left ( \sec ^{2}\theta -1 \right )}}$
$\Rightarrow \frac{2\sec \theta }{\sqrt{\tan ^{2}\theta }}$                 $\left ( \therefore \sec ^{2}\theta = 1+\tan ^{2} \theta \right )$
$\Rightarrow \frac{2\sec \theta }{\tan \theta }$
$\Rightarrow 2\cdot \frac{1}{\cos \theta }\times \frac{\cos \theta }{\sin \theta }$ $\left ( \therefore \frac{1}{\cos \theta } = \sec \theta \right )$
$\Rightarrow \frac{2}{\sin \theta }$
$\Rightarrow 2\, cosec\, \theta$                         $\left ( \therefore \frac{1}{\sin \theta }= cosec\, \theta \right )$
L.H.S = R.H.S
Hence proved