Question

Prove that $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\tan \theta +\cot \theta$.

Answer 1

Consider the given equation.

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

We know that,

$\tan \theta =\frac{\sin \theta }{\cos \theta }$ and $\cot \theta =\frac{\cos \theta }{\sin \theta }$

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Consider the $LHS$.

$\Rightarrow \frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }$

$\Rightarrow \frac{\frac{\sin \theta }{\cos \theta }}{1-\frac{cos\theta }{\sin \theta }}+\frac{\frac{cos\theta }{\sin \theta }}{1-\frac{\sin \theta }{\cos \theta }}$

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

$\Rightarrow \frac{\tan \theta \sin \theta }{(\sin \theta -\cos \theta )}+\frac{\cot \theta \cos \theta }{(\cos \theta -\sin \theta )}$

$\Rightarrow \frac{1}{(\sin \theta -\cos \theta )}[\tan \theta \sin \theta -\cot \theta \cos \theta ]$

$\Rightarrow \frac{1}{(\sin \theta -\cos \theta )}\left[\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta \cos \theta }\right]$

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

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

$\Rightarrow 1+\tan \theta +\cot \theta$

$\Rightarrow RHS$

Hence, proved.