# Solve this

Question:

$(1+\tan \theta+\cot \theta)(\sin \theta-\cos \theta)=\left(\frac{\sec \theta}{\operatorname{cosec}^{2} \theta}-\frac{\operatorname{cosec} \theta}{\sec ^{2} \theta}\right)$

Solution:

$\mathrm{LHS}=(1+\tan \theta+\cot \theta)(\sin \theta-\cos \theta)$

$=\sin \theta+\tan \theta \sin \theta+\cot \theta \sin \theta-\cos \theta-\tan \theta \cos \theta-\cot \theta \cos \theta$

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

$=\sin \theta+\tan \theta \sin \theta+\cos \theta-\cos \theta-\sin \theta-\cot \theta \cos \theta$

$=\tan \theta \sin \theta-\cot \theta \cos \theta$

$=\frac{\sin \theta}{\cos \theta} \times \frac{1}{\operatorname{cosec} \theta}-\frac{\cos \theta}{\sin \theta} \times \frac{1}{\sec \theta}$

$=\frac{1}{\operatorname{cosec} \theta} \times \frac{1}{\operatorname{cosec} \theta} \times \sec \theta-\frac{1}{\sec \theta} \times \frac{1}{\sec \theta} \times \operatorname{cosec} \theta$

$=\frac{\sec \theta}{\operatorname{cosec}^{2} \theta}-\frac{\operatorname{cosec} \theta}{\sec ^{2} \theta}$

$=\mathrm{RHS}$

Hence, LHS $=$ RHS