Question

please answer this question​

Answer 1

$\underline{\textbf{To prove:}}$

$\mathsf{\sqrt{\dfrac{cosecA+2\,cosA}{cosecA-2\,cosA}}=\dfrac{tanA+1}{tanA-1}}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{cosecA+2\,cosA}{cosecA-2\,cosA}}$

$\mathsf{=\dfrac{\dfrac{1}{sinA}+2\,cosA}{\dfrac{1}{sinA}-2\,cosA}}$

$\mathsf{=\dfrac{\dfrac{1+2\,sinA\,cosA}{sinA}}{\dfrac{1-2\,sinA\,cosA}{sinA}}}$

$\mathsf{=\dfrac{1+2\,sinA\,cosA}{1-2\,sinA\,cosA}}$

$\textsf{Using the identity,}$

$\boxed{\mathsf{sin\,2A=2\,sinA\,cosA}}$

$\mathsf{=\dfrac{1+sin\,2A}{1-sin2\,A}}$

$\textsf{Using the identity,}$

$\boxed{\mathsf{sin\,2A=\dfrac{2\,tanA}{1+tan^2A}}}$

$\mathsf{=\dfrac{1+\dfrac{2\,tanA}{1+tan^2A}}{1-\dfrac{2\,tanA}{1+tan^2A}}}$

$\mathsf{=\dfrac{\dfrac{1+tan^2A+2\,tanA}{1+tan^2A}}{\dfrac{1+tan^2A-2\,tanA}{1+tan^2A}}}$

$\mathsf{=\dfrac{\dfrac{(tanA+1)^2}{1+tan^2A}}{\dfrac{(tanA-1)^2}{1+tan^2A}}}$

$\mathsf{=\dfrac{(tanA+1)^2}{(tanA-1)^2}}$

$\mathsf{=\left(\dfrac{tanA+1}{tanA-1}\right)^2}$

$\implies\mathsf{\dfrac{cosecA+2\,cosA}{cosecA-2\,cosA}=\left(\dfrac{tanA+1}{tanA-1}\right)^2}$

$\textsf{Taking square root on botsides, we get}$

$\mathsf{\sqrt{\dfrac{cosecA+2\,cosA}{cosecA-2\,cosA}}=\sqrt{\left(\dfrac{tanA+1}{tanA-1}\right)^2}}$

$\boxed{\mathsf{\sqrt{\dfrac{cosecA+2\,cosA}{cosecA-2\,cosA}}=\dfrac{tanA+1}{tanA-1}}}$