Question

Prove: $(1+\tan \theta +\cot \theta )(\sin \theta -\cos \theta )=\frac{\text{sec}\theta }{{\text{cosec}}^2\theta }-\frac{\text{cosec}\theta }{{\text{sec}}^2\theta }$

Answer 1

Lets take L.H.S and then equate it to R.H.S.

L.H.S$=(1+\tan \theta +\cot \theta )(\sin \theta -\cos \theta )$

$=\sin \theta +\tan \theta \sin \theta +\cot \theta \sin \theta -\cos \theta -\tan \theta \cos \theta -\cot \theta \cos \theta$
$=\sin \theta +\tan \theta \sin \theta +\frac{\cos \theta }{\sin \theta }\times \sin \theta -\cos \theta -\frac{\sin \theta }{\cos \theta }\times \cos \theta -\cot \theta \cos \theta$
$=\sin \theta +\tan \theta \sin \theta +\cos \theta -\cos \theta -\sin \theta -\cot \theta \cos \theta$
$=\tan \theta \sin \theta -\cot \theta \cos \theta$
$=\frac{\sin \theta }{\cos \theta }\times \frac{1}{\text{cosec}\theta }-\frac{\cos \theta }{\sin \theta }\times \frac{1}{\sec \theta }$

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

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

$=\frac{1}{{\text{cosec}}^2\theta }\times \sec \theta -\frac{1}{{\text{sec}}^2\theta }\times \text{cosec}\theta$

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

$=R.H.S$

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