Question

Prove the following:
\[\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta \cdot \ co sec\theta \]

Answer 1

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,
\[\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta \cdot \ co sec\theta \]

Taking LHS, we get
\[\dfrac{{\tan \theta }}{{1 - \cot \theta }} + \dfrac{{\cot \theta }}{{1 - \tan \theta }}\]

Now, we will make it in terms of \[\sin \theta \] and $\cos \theta $.
We know that, \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] and \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\], then it becomes,
= \[\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{1 - \dfrac{{\cos \theta }}{{\sin \theta }}}} + \dfrac{{\dfrac{{\cos \theta }}{{\sin \theta }}}}{{1 - \dfrac{{\sin \theta }}{{\cos \theta }}}}\]
Now, by taking LCM in the denominator, we get

=\[\dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{\dfrac{{\sin \theta - \cos \theta }}{{\sin \theta }}}} + \dfrac{{\dfrac{{\cos \theta }}{{\sin }}}}{{\dfrac{{\cos \theta - \sin \theta }}{{\cos \theta }}}}\]
= \[\dfrac{{\sin \theta }}{{\cos \theta }} \times \dfrac{{\sin \theta }}{{\sin \theta - \cos \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }} \times \dfrac{{\cos \theta }}{{\cos \theta - \sin \theta }}\]

= \[\dfrac{{{{\sin }^2}\theta }}{{\cos \theta \left( {\sin \theta - \cos \theta } \right)}} + \dfrac{{{{\cos }^2}\theta }}{{\sin \theta \left( {\cos \theta - \sin \theta } \right)}}\]
By taking minus common from the second term,
= \[\dfrac{{{{\sin }^2}\theta }}{{\cos \theta \left( {\sin \theta - \cos \theta } \right)}} - \dfrac{{{{\cos }^2}\theta }}{{\sin \theta \left( {\sin \theta - \cos \theta } \right)}}\]

= \[\dfrac{{{{\sin }^3}\theta - {{\cos }^3}\theta }}{{\cos \theta \cdot \sin \theta \left( {\sin \theta - \cos \theta } \right)}}\]
Now, in the numerator, \[{\sin ^3}\theta - {\cos ^3}\theta \] represents an identity, i.e., \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\], therefore, it becomes,
= \[\dfrac{{\left( {\sin \theta - \cos \theta } \right)\left( {{{\sin }^2}\theta + \sin \theta \cdot \cos \theta + {{\cos }^2}\theta } \right)}}{{\cos \theta \cdot \sin \theta \left( {\sin \theta - \cos \theta } \right)}}\]
So, by cancelling the equal terms in the numerator and denominator, we obtain
=\[\dfrac{{\left( {{{\sin }^2}\theta + {{\cos }^2}\theta + \sin \theta \cdot \cos \theta } \right)}}{{\cos \theta \cdot \sin \theta }}\]
= \[\dfrac{{1 + \sin \theta \cdot \cos \theta }}{{\cos \theta \cdot \sin \theta }}\] \[\left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]\]
Now, by separating the terms, we obtain
= \[\dfrac{1}{{\cos \theta \cdot \sin \theta }} + \dfrac{{\sin \theta \cdot \cos \theta }}{{sin\theta \cdot \cos \theta }}\]
By cancelling out the equal terms,
= \[\sec \theta \cdot \ co sec\theta + 1\] , which can also be written as,
=\[1 + \sec \theta \cdot \ co sec\theta \]
=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.