1. ## trig identity

I spent 1 hour just trying to solve this, but I don't get it
need some help, thanks

2. Hello, livmed!

"Solve" is the wrong word.
You want to prove the identity.

$
\left(\frac{\sin x + \tan x}{\dfrac{1}{\sin x} + \cot x}\right)^2 \;=\; \frac{\sin^2\!x + \tan^2\!x}{\dfrac{1}{\sin^2\!x} + \cot^2\!x}$

The left side: . $\left(\frac{\sin x + \tan x}{\dfrac{1}{\sin x} + \cot x}\right)^2 \;=\;\left(\frac{\sin x + \dfrac{\sin x}{\cos x}}{\dfrac{1}{\sin x} + \dfrac{\cos x}{\sin x}}\right)^2$

. . . . . . $=\; \left(\frac{\sin x\left(1 + \dfrac{1}{\cos x}\right)} {\dfrac{1+\cos x}{\sin x}}\right)^2 \;=\;\left(\frac{\sin x\left(\dfrac{\cos x + 1}{\cos x}\right)}{\dfrac{1+\cos x}{\sin x}}\right)^2$ . $=\; \left(\frac{\sin^2\!x}{\cos x}\right)^2 \;=\;\frac{\sin^4\!x}{\cos^2\!x}$

$\text{Multiply by }\frac{1+\cos^2\!x}{1+\cos^2\!x}\!:\quad\frac{\sin ^2\!x\cdot\sin^2\!x}{\cos^2\!x}\cdot\frac{1+\cos^2 \!x}{1 + \cos^2\!x} \;=\;\frac{\sin^2\!x}{\cos^2\!x}\cdot\frac{1+\cos^ 2\!x}{\dfrac{1+\cos^2\!x}{\sin^2\!x}}$

. . . . . . $=\; \frac{\tan^2\!x(1 + \cos^2\!x)}{\dfrac{1+\cos^2\!x}{\sin^2\!x}} \;=\;\frac{\tan^2\!x + \tan^2\!x\cos^2\!x}{\dfrac{1}{\sin^2\!x} + \dfrac{\cos^2\!x}{\sin^2\!x}}$

. . . . . . $= \;\frac{\tan^2\!x + \sin^2\!x}{\dfrac{1}{\sin^2\!x} + \cot^2\!x}$