Question

sqrt[(cosecx-1)/(cosec+1)]=1/(secx+tanx)

Answer 1

here is ur answer
.. hope this helps u

Answer 2

$\sqrt{\dfrac{\csc x-1}{\csc x+1}}=\dfrac{1}{\sec x +\tan x}$, proved.

Step-by-step explanation:

L.H.S. =$\sqrt{\dfrac{\csc x-1}{\csc x+1}}$

Multiplying numerator and denominator by $\csc x+1$, we het

$=\sqrt{\dfrac{\csc x-1}{\csc x+1}\times \dfrac{\csc x-1}{\csc x-1}}$

$=\sqrt{\dfrac{(\csc x-1)^{2} }{\csc^2 x-1}$

$=\sqrt{\dfrac{(\csc x-1)^{2} }{\cot ^2x}$

$=\dfrac{(\csc x-1)}{\cot x}$

$=\sec x -\tan x$

$=\dfrac{(\sec x -\tan x)(\sec x +\tan x)}{\sec x +\tan x}$

$=\dfrac{(\sec^{2} x -\tan^{2} x)}{\sec x +\tan x}$

$=\dfrac{1}{\sec x +\tan x}$ [∵ $\sec^{2} A -\tan^{2} A=1$]

= R.H.S, proved