Prove that,$(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)=\sec \alpha+\operatorname{cosec} \alpha$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
$LHS$ $= (\sin \alpha + \cos \alpha)(\tan \alpha + \cot \alpha)$
$= (\sin \alpha + \cos \alpha) \left( \frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} \right)$ $\left[ \because \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \text{ and } \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \right]$
$= (\sin \alpha + \cos \alpha) \left( \frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha} \right)$
$= (\sin \alpha + \cos \alpha) \cdot \frac{1}{\sin \alpha \cos \alpha}$ $\left[ \because \sin^2 \alpha + \cos^2 \alpha = 1 \right]$
$= \frac{\sin \alpha}{\sin \alpha \cos \alpha} + \frac{\cos \alpha}{\sin \alpha \cos \alpha}$
$= \frac{1}{\cos \alpha} + \frac{1}{\sin \alpha}$
$= \sec \alpha + \operatorname{cosec} \alpha = \text{RHS}$

Explore More

Similar Questions

$\tan (65^\circ - \theta) - \cot (25^\circ + \theta) - \sec (55^\circ - \theta) + \operatorname{cosec}(35^\circ + \theta) = \ldots \ldots \ldots \ldots$ (where,$0 < \theta < 25^\circ$)

Prove the following identity,where the angle involved is an acute angle for which the expression is defined:
$\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$

$\frac{\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}=\ldots \ldots \ldots \ldots$

Prove that $\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}=\tan \theta+\cot \theta$

Difficult
View Solution

If $a \sin \theta + b \cos \theta = c$,then prove that $a \cos \theta - b \sin \theta = \pm \sqrt{a^2 + b^2 - c^2}$,given $a^2 + b^2 \geq c^2$.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo