Prove that,
$1 + \frac{\cot^{2} \alpha}{1 + \operatorname{cosec} \alpha} = \operatorname{cosec} \alpha$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) $L$.$H$.$S$. $= 1 + \frac{\cot^{2} \alpha}{1 + \operatorname{cosec} \alpha}$
$= 1 + \frac{\cos^{2} \alpha / \sin^{2} \alpha}{1 + 1 / \sin \alpha} \quad [\because \cot \theta = \frac{\cos \theta}{\sin \theta} \text{ and } \operatorname{cosec} \theta = \frac{1}{\sin \theta}]$
$= 1 + \frac{\cos^{2} \alpha}{\sin \alpha(1 + \sin \alpha)} = \frac{\sin \alpha(1 + \sin \alpha) + \cos^{2} \alpha}{\sin \alpha(1 + \sin \alpha)}$
$= \frac{\sin \alpha + (\sin^{2} \alpha + \cos^{2} \alpha)}{\sin \alpha(1 + \sin \alpha)} \quad [\because \sin^{2} \theta + \cos^{2} \theta = 1]$
$= \frac{\sin \alpha + 1}{\sin \alpha(1 + \sin \alpha)} = \frac{1}{\sin \alpha} \quad [\because \operatorname{cosec} \theta = \frac{1}{\sin \theta}]$
$= \operatorname{cosec} \alpha = \text{R.H.S.}$

Explore More

Similar Questions

$(1+\tan ^{2} \theta)(1-\cos ^{2} \theta) = \dots$

If $\tan \theta + \sec \theta = l$,then prove that $\sec \theta = \frac{l^{2} + 1}{2l}$.

The value of the expression $\left[\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right]$ is:

Difficult
View Solution

In $\Delta ABC$,$m \angle C = 90^{\circ}$ and $\tan A = \frac{1}{\sqrt{3}}$,then $\sin A = \ldots$

If $\sec \theta = \frac{13}{5}$,then $\cos \theta = \ldots \ldots \ldots \ldots$

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