સાબિત કરો કે,
$\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \operatorname{cosec} \theta$

(N/A) ડા.બા. $= \frac{\sin \theta}{1+\cos \theta} + \frac{1+\cos \theta}{\sin \theta}$
$= \frac{\sin^2 \theta + (1+\cos \theta)^2}{\sin \theta(1+\cos \theta)}$
$= \frac{\sin^2 \theta + 1 + \cos^2 \theta + 2 \cos \theta}{\sin \theta(1+\cos \theta)}$ $\quad [\because (a+b)^2 = a^2 + b^2 + 2ab]$
$= \frac{(\sin^2 \theta + \cos^2 \theta) + 1 + 2 \cos \theta}{\sin \theta(1+\cos \theta)}$
$= \frac{1 + 1 + 2 \cos \theta}{\sin \theta(1+\cos \theta)}$ $\quad [\because \sin^2 \theta + \cos^2 \theta = 1]$
$= \frac{2 + 2 \cos \theta}{\sin \theta(1+\cos \theta)}$
$= \frac{2(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$
$= \frac{2}{\sin \theta} = 2 \operatorname{cosec} \theta = \text{જ.બા.}$ $\quad [\because \operatorname{cosec} \theta = \frac{1}{\sin \theta}]$