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Question: Prove the following: \[\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \t...

Prove the following:
tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θ⋅ cosecθ\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta \cdot \ co sec\theta

Explanation

Solution

Hint - We will prove that the Left Hand Side and the Right Hand Side of the given trigonometric equation are equal, for which, we first write down the given equation and start solving either the LHS or the RHS of the equation by taking the help of various trigonometric identities.

Complete step-by-step answer:
The given equation in the question is,
tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θ⋅ cosecθ\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta \cdot \ co sec\theta

Taking LHS, we get
tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }}

Now, we will make it in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta .
We know that, tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} and cot⁡θ=cos⁡θsin⁡θ\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}, then it becomes,
= sin⁡θcos⁡θ1−cos⁡θsin⁡θ+cos⁡θsin⁡θ1−sin⁡θcos⁡θ\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{1 - \dfrac{{\cos \theta }}{{\sin \theta }}}} + \dfrac{{\dfrac{{\cos \theta }}{{\sin \theta }}}}{{1 - \dfrac{{\sin \theta }}{{\cos \theta }}}}
Now, by taking LCM in the denominator, we get

=sin⁡θcos⁡θsin⁡θ−cos⁡θsin⁡θ+cos⁡θsin⁡cos⁡θ−sin⁡θcos⁡θ\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{\dfrac{{\sin \theta - \cos \theta }}{{\sin \theta }}}} + \dfrac{{\dfrac{{\cos \theta }}{{\sin }}}}{{\dfrac{{\cos \theta - \sin \theta }}{{\cos \theta }}}}
= sin⁡θcos⁡θ×sin⁡θsin⁡θ−cos⁡θ+cos⁡θsin⁡θ×cos⁡θcos⁡θ−sin⁡θ\dfrac{{\sin \theta }}{{\cos \theta }} \times \dfrac{{\sin \theta }}{{\sin \theta - \cos \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }} \times \dfrac{{\cos \theta }}{{\cos \theta - \sin \theta }}

= sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ)+cos⁡2θsin⁡θ(cos⁡θ−sin⁡θ)\dfrac{{{{\sin }^2}\theta }}{{\cos \theta \left( {\sin \theta - \cos \theta } \right)}} + \dfrac{{{{\cos }^2}\theta }}{{\sin \theta \left( {\cos \theta - \sin \theta } \right)}}
By taking minus common from the second term,
= sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ)−cos⁡2θsin⁡θ(sin⁡θ−cos⁡θ)\dfrac{{{{\sin }^2}\theta }}{{\cos \theta \left( {\sin \theta - \cos \theta } \right)}} - \dfrac{{{{\cos }^2}\theta }}{{\sin \theta \left( {\sin \theta - \cos \theta } \right)}}

= sin⁡3θ−cos⁡3θcos⁡θ⋅sin⁡θ(sin⁡θ−cos⁡θ)\dfrac{{{{\sin }^3}\theta - {{\cos }^3}\theta }}{{\cos \theta \cdot \sin \theta \left( {\sin \theta - \cos \theta } \right)}}
Now, in the numerator, sin⁡3θ−cos⁡3θ{\sin ^3}\theta - {\cos ^3}\theta represents an identity, i.e., a3−b3=(a−b)(a2+ab+b2){a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right), therefore, it becomes,
= (sin⁡θ−cos⁡θ)(sin⁡2θ+sin⁡θ⋅cos⁡θ+cos⁡2θ)cos⁡θ⋅sin⁡θ(sin⁡θ−cos⁡θ)\dfrac{{\left( {\sin \theta - \cos \theta } \right)\left( {{{\sin }^2}\theta + \sin \theta \cdot \cos \theta + {{\cos }^2}\theta } \right)}}{{\cos \theta \cdot \sin \theta \left( {\sin \theta - \cos \theta } \right)}}
So, by cancelling the equal terms in the numerator and denominator, we obtain
=(sin⁡2θ+cos⁡2θ+sin⁡θ⋅cos⁡θ)cos⁡θ⋅sin⁡θ\dfrac{{\left( {{{\sin }^2}\theta + {{\cos }^2}\theta + \sin \theta \cdot \cos \theta } \right)}}{{\cos \theta \cdot \sin \theta }}
= 1+sin⁡θ⋅cos⁡θcos⁡θ⋅sin⁡θ\dfrac{{1 + \sin \theta \cdot \cos \theta }}{{\cos \theta \cdot \sin \theta }} [∵sin⁡2θ+cos⁡2θ=1]\left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]
Now, by separating the terms, we obtain
= 1cos⁡θ⋅sin⁡θ+sin⁡θ⋅cos⁡θsinθ⋅cos⁡θ\dfrac{1}{{\cos \theta \cdot \sin \theta }} + \dfrac{{\sin \theta \cdot \cos \theta }}{{sin\theta \cdot \cos \theta }}
By cancelling out the equal terms,
= sec⁡θ⋅ cosecθ+1\sec \theta \cdot \ co sec\theta + 1 , which can also be written as,
=1+sec⁡θ⋅ cosecθ1 + \sec \theta \cdot \ co sec\theta
=RHS
Hence Proved.

Note – One should be careful while doing such questions because in these questions we have to use different identities which might be confusing sometimes. So, there should be no doubt about any identity and before doing these questions one must be clear in all aspects.