Question

Last one ^_^

\(\sf\dfrac{sinx}{1-cosx}+\dfrac{sinx}{1+cosx}=2cscx\)

Answer 1

\(\large\tt \begin{align} \color{black}{

\dfrac{\sin x}{1-\cos x}+\dfrac{\ sin x}{1+\cos x}\\~\\
=\sin x\left(\dfrac{1}{1-\cos x}+\dfrac{1}{1+\cos x}\right)\\~\\
=\sin x\left(\dfrac{1+\cos x+1-\cos x}{1-\cos^2 x}\right)\\~\\
=\sin x\left(\dfrac{2}{1-\cos^2 x}\right)\\~\\
=\sin x\left(\dfrac{2}{sin^2 x}\right)\\~\\
}\end{align}\)

Answer 2

@Brostep0s mathmath333's solution is correct.

Answer 3

thanks @ParthKohli :)