Question

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

Answer 1

Simplify the LHS of $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\tan \theta +\cot \theta$

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

$=\frac{{\tan }^2\theta }{\tan \theta -1}+\frac{1}{\tan \theta \left(1-\tan \theta \right)}$

$=\frac{{\tan }^3\theta -1}{\tan \theta \left(\tan \theta -1\right)}$

$=\frac{\left(\tan \theta -1\right)\left({\tan }^2\theta +\tan \theta +1\right)}{\tan \theta \left(\tan \theta -1\right)}$

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

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

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

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

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

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

$=1+\sec \theta \csc \theta$

Now simplify the RHS of $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\tan \theta +\cot \theta$

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

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

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

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

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

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

$=1+\sec \theta \csc \theta$

Therefore, $LHS=RHS=1+\sec \theta \csc \theta$.