Question

please answer correct or I will report please answer

Answer 1

Step-by-step explanation:

Note: I am replacing θ with A for the better understanding.

$Given:\frac{sinA}{1-cosA} + \frac{tanA}{1+cosA}$

$=\frac{sinA}{1-cosA}*\frac{1+cosA}{1+cosA}+\frac{tanA}{1+cosA}*\frac{1-cosA}{1-cosA}$

$=\frac{sinA(1+cosA)}{(1-cosA)(1+cosA)}+\frac{tanA(1-cosA)}{(1+cosA)(1-cosA)}$

$=\frac{sinA(1+cosA)}{1-cos^2A}+ \frac{tanA(1-cosA)}{1-cos^2A}$

$=\frac{sinA(1+cosA)}{sin^2A}+\frac{tanA(1-cosA)}{sin^2A}$

$=\frac{sinA(1+cosA)}{sin^2A} + \frac{sinA(1-cosA)}{cosAsin^2A}$

$=sinA[\frac{1+cosA}{sin^2A}+\frac{1-cosA}{cosAsin^2A}]$

$=\frac{1+cosA}{sinA}+\frac{1-cosA}{cosAsinA}$

$=\frac{1}{sinA}+\frac{cosA}{sinA} + \frac{1}{cosAsinA} - \frac{cosA}{sinAcosA}$

$=cosecA + cotA - secAcosecA - cosecA$

$=secAcosecA + cotA$

Hope it helps!

Answer 2

The answer is explained below.