Question

Prove that $\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta \operatorname{cosec} \theta $

Answer 1

Hint: Convert LHS into the terms of $\tan \theta $ and simplify using the formula for ${a^3} - {b^3}$, change to the trigonometric functions on RHS to get the desired result.

Complete step-by-step answer:

Let us consider the L.H.S. and convert $\cot \theta $ into $\tan \theta $
LHS=
$\dfrac{{\tan \theta }}{{1 - \dfrac{1}{{\tan \theta }}}} + \dfrac{{\dfrac{1}{{\tan \theta }}}}{{1 - \tan \theta }}$
= $\dfrac{{\tan \theta }}{{\dfrac{{\tan \theta - 1}}{{\tan \theta }}}} - \dfrac{{\dfrac{1}{{\tan \theta }}}}{{\tan \theta - 1}}$
= $\dfrac{{{{\tan }^2}\theta }}{{\tan \theta - 1}} - \dfrac{{\dfrac{1}{{\tan \theta }}}}{{\tan \theta - 1}}$
= $\dfrac{{{{\tan }^3}\theta - 1}}{{\tan \theta \left( {\tan \theta - 1} \right)}}$

We know that ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$, so we get,
= $\dfrac{{\left( {\tan \theta - 1} \right)\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta \left( {\tan \theta - 1} \right)}}$
= $\dfrac{{{{\tan }^2}\theta + \tan \theta + 1}}{{\tan \theta }}$
= $\tan \theta + 1 + \cot \theta $
= $\dfrac{{\sin \theta }}{{\cos \theta }} + 1 + \dfrac{{\cos \theta }}{{\sin \theta }}$
= $1 + \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\sin \theta cos\theta }}$ $\left( {{{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right)$
= $1 + \sec \theta \operatorname{cosec} \theta $ = RHS

Hence Proved.

Note: In these questions it is advised to simplify the LHS or the RHS according to their complexity of trigonometric functions. Sometimes proving LHS = RHS needs simplification on both sides of the equation. Remember to convert dissimilar trigonometric functions to get to the final result.