Prove that:$ \frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}=\cot \theta $

To do:

We have to prove that \( \frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}=\cot \theta \).

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

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

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

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

$=\frac{\cos \theta(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$

$=\frac{\cos \theta}{\sin \theta}$

$=\cot \theta$

Hence proved.