Question

Someone please help me with this question

Answer 1

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 \thetasecθ+1secθ−1+secθ−1secθ+1=2cscθ

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

First let’s solve LHS:

L.H.S.   \begin{gathered}\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}\end{gathered}→secθ+1secθ−1+secθ−1secθ+1=(secθ+1)(secθ−1)(secθ−1)(secθ−1)+(secθ−1)(secθ+1)(secθ+1)(secθ+1)=sec2θ−1(secθ−1)2+sec2θ−1(secθ+1)2=tan2θ(secθ−1)2+tan2θ(secθ+1)2

[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}1+tan2θ=sec2θ;∴sec2θ−1=tan2θ]=tanθsecθ−1+tanθsecθ+1 [Square rooting, since \bold{\sqrt{a^{2}}=a}a2=a

\begin{gathered}\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}\end{gathered}=tanθsecθ−1+secθ+1=tanθ2secθ=2tanθsecθ=2cosθsinθ1

[Since, \sec \theta=\frac{1}{\cos \theta} \quad \text { and } \tan \theta=\frac{\sin \theta}{\cos \theta}secθ=cosθ1 and tanθ=cosθsinθ ]

\begin{gathered}\begin{array}{l}{=2 \times \frac{1}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}} \\ {=2 \times \frac{1}{\sin \theta}}\end{array}\end{gathered}=2×cosθ1×sinθcosθ=2×sinθ1

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