Question

$\sqrt{\frac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\frac{\sec \theta +1}{\sec \theta -1}}$= ______.

• A. $2\csc \theta$
• B. $\frac{2\sin \theta }{\sqrt{\sec \theta }}$
• C. $2\cos \theta$
• D. $2\sec \theta$

Answer 1

The correct option is A $2\csc \theta$

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

$LHS=\sqrt{\frac{(\sec \theta -1)(\sec \theta -12)}{(\sec \theta +1)(\sec \theta -1)}}+\sqrt{\frac{(\sec \theta +1)(\sec \theta +1)}{(\sec \theta -1)(\sec \theta +1)}}$

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

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

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

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

$=\frac{2\sec \theta }{\tan \theta }$

$=2\times \frac{1}{\cos \theta }\times \frac{\cos \theta }{\sin \theta }$

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

$=2cosec\theta$

$=R.H.S$