Question

Prove: under root (sec theta-1/sec theta+1) + under root (sec theta+1/sec theta-1) = 2 Cosec theta?

Answer 1

Hope this helps.Its done step by step and easy to understand.The denominator is rationalized in the first step...

Answer 2

Answer:

$\sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}=2 \csc \theta$ is proved.

Step-by-step explanation:

Given identity to prove: $\sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}=2 \csc \theta$

For convenience in proving it is taken as LHS and RHS.

First let’s solve LHS:

L.H.S.   $\rightarrow \sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}\begin{array}{l}{=\sqrt{\frac{(\sec \theta-1)(\sec \theta-1)}{(\sec \theta+1)(\sec \theta-1)}}+\sqrt{\frac{(\sec \theta+1)(\sec \theta+1)}{(\sec \theta-1)(\sec \theta+1)}}} \\ {=\sqrt{\frac{(\sec \theta-1)^{2}}{\sec ^{2} \theta-1}}+\sqrt{\frac{(\sec \theta+1)^{2}}{\sec ^{2} \theta-1}}=\sqrt{\frac{(\sec \theta-1)^{2}}{\tan ^{2} \theta}}+\sqrt{\frac{(\sec \theta+1)^{2}}{\tan ^{2} \theta}}}\end{array}$

[Since, $1+\tan ^{2} \theta=\sec ^{2} \theta ; \quad \therefore \sec ^{2} \theta-1=\tan ^{2} \theta]=\frac{\sec \theta-1}{\tan \theta}+\frac{\sec \theta+1}{\tan \theta}$ [Square rooting, since $\bold{\sqrt{a^{2}}=a}$]

$\begin{array}{l}{=\frac{\sec \theta-1+\sec \theta+1}{\tan \theta}} \\ {=\frac{2 \sec \theta}{\tan \theta}=2 \frac{\sec \theta}{\tan \theta}} \\ {=2 \frac{\frac{1}{\sin \theta}}{\cos \theta}}\end{array}$

[Since, $\sec \theta=\frac{1}{\cos \theta} \quad \text { and } \tan \theta=\frac{\sin \theta}{\cos \theta}$ ]

$\begin{array}{l}{=2 \times \frac{1}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}} \\ {=2 \times \frac{1}{\sin \theta}}\end{array}$

=2 cosecθ     [Since,1/sin⁡θ =cos⁡θ]  = R.H.S.    

 

Thus LHS = RHS.  

Hence proved.