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Question: Prove that \(\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }}...

Prove that tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θcosec⁡θ\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta \operatorname{cosec} \theta

Explanation

Solution

Hint: Convert LHS into the terms of tan⁡θ\tan \theta and simplify using the formula for a3−b3{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⁡θ\cot \theta into tan⁡θ\tan \theta
LHS=
tan⁡θ1−1tan⁡θ+1tan⁡θ1−tan⁡θ\dfrac{{\tan \theta }}{{1 - \dfrac{1}{{\tan \theta }}}} + \dfrac{{\dfrac{1}{{\tan \theta }}}}{{1 - \tan \theta }}
= tan⁡θtan⁡θ−1tan⁡θ−1tan⁡θtan⁡θ−1\dfrac{{\tan \theta }}{{\dfrac{{\tan \theta - 1}}{{\tan \theta }}}} - \dfrac{{\dfrac{1}{{\tan \theta }}}}{{\tan \theta - 1}}
= tan⁡2θtan⁡θ−1−1tan⁡θtan⁡θ−1\dfrac{{{{\tan }^2}\theta }}{{\tan \theta - 1}} - \dfrac{{\dfrac{1}{{\tan \theta }}}}{{\tan \theta - 1}}
= tan⁡3θ−1tan⁡θ(tan⁡θ−1)\dfrac{{{{\tan }^3}\theta - 1}}{{\tan \theta \left( {\tan \theta - 1} \right)}}

We know that a3−b3=(a−b)(a2+b2+ab){a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right), so we get,
= (tan⁡θ−1)(tan⁡2θ+1+tan⁡θ)tan⁡θ(tan⁡θ−1)\dfrac{{\left( {\tan \theta - 1} \right)\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta \left( {\tan \theta - 1} \right)}}
= tan⁡2θ+tan⁡θ+1tan⁡θ\dfrac{{{{\tan }^2}\theta + \tan \theta + 1}}{{\tan \theta }}
= tan⁡θ+1+cot⁡θ\tan \theta + 1 + \cot \theta
= sin⁡θcos⁡θ+1+cos⁡θsin⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} + 1 + \dfrac{{\cos \theta }}{{\sin \theta }}
= 1+sin⁡2θ+cos⁡2θsin⁡θcosθ1 + \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\sin \theta cos\theta }} (sin⁡2θ+cos⁡2θ=1)\left( {{{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right)
= 1+sec⁡θcosec⁡θ1 + \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.