8. Introduction to Trigonometry
medium

निम्नलिखित को सिद्ध कीजिए :

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

Option A
Option B
Option C
Option D

Solution

L.H.S.$=1+\frac{\cot ^{2} \alpha}{1+\operatorname{cosec} \alpha}=1+\frac{\cos ^{2} \alpha / \sin ^{2} \alpha}{1+1 / \sin \alpha} \quad\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right.$ and $\left.\operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$

$=1+\frac{\cos ^{2} \alpha}{\sin \alpha(1+\sin \alpha)}=\frac{\sin \alpha(1+\sin \alpha)+\cos ^{2} \alpha}{\sin \alpha(1+\sin \alpha)}$

$=\frac{\sin \alpha+\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)}{\sin \alpha(1+\sin \alpha)}$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$=\frac{(\sin \alpha+1)}{\sin \alpha(\sin \alpha+1)}=\frac{1}{\sin \alpha}$ $\left[\because \operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$

$=\operatorname{cosec} \alpha=$ R.H.S.

Standard 10
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.