Question

very difficult for me
please answer

Answer 1

$Given LHS = \sqrt{ \frac{cosecx - 1}{cosex + 1} }$

On rationalizing, we get

$\sqrt{ \frac{(cosecx - 1)(cosecx-1)}{(cosecx+1)(cosecx-1)} }$

$\sqrt{ \frac{(cosecx-1)^2}{cosec^2x - 1} }$

$\sqrt{ \frac{(cosecx-1)^2}{cot^2x} }$

$\frac{cosecx - 1}{cotx}$

$\frac{cosecx}{cotx} - \frac{1}{cotx}$

$\frac{ \frac{1}{sinx}}{ \frac{cosx}{sinx} } - tanx$

$\frac{1}{cosx} - tan x$

sec x - tan x

Divide the numerator & denominator with (sec x + tan x)

$\frac{(secx-tanx)(secx+tanx)}{secx+tanx}$

$\frac{(sec^2x - tan^2x)}{secx+tanx}$

We know that sec^2x - tan^2x = 1.

$\frac{1}{secx + tanx}$


LHS = RHS.


Hope this helps!

Answer 2

Maybe this one is the simplest one.