सिद्ध कीजिए कि $\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}=\tan \theta+\cot \theta$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) $L.H.S. = \sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}$
$= \sqrt{\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}}$ $\left[\because \sec \theta = \frac{1}{\cos \theta} \text{ और } \operatorname{cosec} \theta = \frac{1}{\sin \theta}\right]$
$= \sqrt{\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cdot \cos ^{2} \theta}} = \sqrt{\frac{1}{\sin ^{2} \theta \cdot \cos ^{2} \theta}}$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta = 1\right]$
$= \frac{1}{\sin \theta \cdot \cos \theta} = \frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cdot \cos \theta}$ $\left[\because 1 = \sin ^{2} \theta+\cos ^{2} \theta\right]$
$= \frac{\sin ^{2} \theta}{\sin \theta \cdot \cos \theta} + \frac{\cos ^{2} \theta}{\sin \theta \cdot \cos \theta}$
$= \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$ $\left[\because \tan \theta = \frac{\sin \theta}{\cos \theta} \text{ और } \cot \theta = \frac{\cos \theta}{\sin \theta}\right]$
$= \tan \theta + \cot \theta = R.H.S.$

Explore More

Similar Questions

यदि $0 < \theta < 90$ और $\sin \theta = \cos 30$ है,तो $2 \tan^2 \theta - 1 = \dots$

यदि $\sin \theta + \sin^2 \theta = 1$ है,तो $\cos^2 \theta + \cos^4 \theta = \dots$

Difficult
View Solution

यदि $\tan 7 \theta \cdot \tan 3 \theta = 1$ है,तो $\theta$ का मान .......... है।

$(1-\cos \theta)(1+\cos \theta) = \dots$

$\sin^{2} 30^{\circ} - \tan 45^{\circ} + \cos^{2} 60^{\circ} - \cot 90^{\circ}$ का मान ........ है।

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