સાબિત કરો કે,$(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)=\sec \alpha+\operatorname{cosec} \alpha$

ડા.બા. $= (\sin \alpha + \cos \alpha)(\tan \alpha + \cot \alpha)$
$= (\sin \alpha + \cos \alpha) \left( \frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} \right)$ $\left[ \because \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \text{ અને } \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \right]$
$= (\sin \alpha + \cos \alpha) \left( \frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha} \right)$
$= (\sin \alpha + \cos \alpha) \cdot \frac{1}{\sin \alpha \cos \alpha}$ $\left[ \because \sin^2 \alpha + \cos^2 \alpha = 1 \right]$
$= \frac{\sin \alpha}{\sin \alpha \cos \alpha} + \frac{\cos \alpha}{\sin \alpha \cos \alpha}$
$= \frac{1}{\cos \alpha} + \frac{1}{\sin \alpha}$
$= \sec \alpha + \operatorname{cosec} \alpha = \text{જ.બા.}$