Question

Find the value of: $\frac{\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }}{1+\sec \theta \csc \theta }$

Answer 1

Given expression : $\frac{\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }}{1+\sec \theta \csc \theta }\dots \left(1\right)$

Consider the numerator

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

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

$=\frac{{\tan }^2\theta }{\tan \theta -1}-\frac{\frac{1}{\tan \theta }}{\tan \theta -1}$

$=\frac{{\tan }^2\theta -\frac{1}{\tan \theta }}{\tan \theta -1}$

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

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

$=\frac{{\tan }^2\theta +\tan \theta +1}{\tan \theta }$

$=\tan \theta +\cos \theta +1$

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

$=\frac{{\sin }^2\theta +{\cos }^2\theta }{\cos \theta \sin \theta }+1=1+\frac{1}{\cos \theta \sin \theta }$

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

Substituting in $\left(1\right)$

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