Question

Sin theta/1+cot theta + 1+costheta/sintheta=2cosec thta

Answer 1

$\frac{sin\theta}{1+cos\theta}+\frac{1+cos\theta}{sin\theta}$

$=\frac{sin^2\theta+(1+cos\theta)^2}{(1+cos\theta)sin\theta}$

Using

$\boxed{\bf\,(a+b)^2=a^2+b^2+2ab}$

$=\frac{sin^2\theta+1^2+cos^2\theta+2\,cos\theta}{(1+cos\theta)sin\theta}$

$=\frac{sin^2\theta+cos^2\theta+,1+2\,cos\theta}{(1+cos\theta)sin\theta}$

Using,

$\boxed{\bf\,sin^2A+cos^2A=1}$

$=\frac{1+1+2\,cos\theta}{(1+cos\theta)sin\theta}$

$=\frac{2+2\,cos\theta}{(1+cos\theta)sin\theta}$

$=\frac{2(1+cos\theta)}{(1+cos\theta)sin\theta}$

$=\frac{2}{sin\theta}$

$=2(\frac{1}{sin\theta})$

$=2\:cosec\theta$

$\implies\boxed{\bf\frac{sin\theta}{1+cos\theta}+\frac{1+cos\theta}{sin\theta}=2\:cosec\theta}$