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

To do:

We have to prove that \( \sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}=2 \operatorname{cosec} \theta \).

Solution:

We know that,

$\sin ^{2} A+\cos^2 A=1$.......(i)

$\operatorname{cosec} A=\frac{1}{\sin A}$......(ii)

Therefore,

$\sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}=\sqrt{\frac{(\sec \theta -1)}{(\sec \theta +1)} \times \frac{(\sec \theta +1)}{(\sec \theta +1)}} +\sqrt{\frac{(\sec \theta +1)}{(\sec \theta -1)} \times \frac{(\sec \theta -1)}{(\sec \theta -1)}}$

$=\sqrt{\frac{\left(\sec^{2} \theta -1\right)}{(\sec \theta +1)^{2}}} +\sqrt{\frac{\left(\sec^{2} \theta -1\right)}{(\sec \theta -1)^{2}}}$

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

$=\frac{\tan \theta }{\sec \theta +1} +\frac{\tan \theta }{\sec \theta -1}$

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

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

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

$=\frac{\sin \theta ( 1-\cos \theta ) +\sin \theta ( 1+\cos \theta )}{1^{2} -\cos^{2} \theta }$

$=\frac{\sin \theta [ 1-\cos \theta +1+\cos \theta ]}{\sin^{2} \theta }$

$=\frac{2}{\sin \theta }$

$=2\operatorname{cosec} \theta$ 

Hence proved.    

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Updated on: 10-Oct-2022

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