# Prove that :

Question:

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+s \operatorname{ec} \theta \operatorname{cosec} \theta$

Solution:

Here we have to prove that

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$

Here we take the LHS and by using the trigonometric identities, we have

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$

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

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

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

$=\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta \cos \theta(\sin \theta-\cos \theta)}$

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

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

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

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

$=1+\sec \theta \operatorname{cosec} \theta$

Hence $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$