Question

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

Answer 1

LHS

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

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

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

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

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

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

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

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

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

$=tan\theta +1+cot\theta$

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

$=1+\frac{si{n}^2\theta +co{s}^2\theta }{sin\theta cos\theta }$

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

$=1+sec\theta cosec\theta$ = RHS