Question

check if $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta$

• A. True
• B. False

Answer 1

The correct option is A True

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

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

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

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

$=\frac{(\sin \theta -\cos \theta )({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta )}{(\sin \theta -\cos \theta )\sin \theta \cos \theta }$ $(\because a^3-b^3=(a-b\left)\right(a^2+b^2+ab\left)\right)$

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

So the relation is $\text{True}$