Question

Prove the identity \[\dfrac{{\sin \theta }}{{1 - \cos \theta }} - \dfrac{1}{{\sin \theta }} = \dfrac{1}{{\tan \theta }}\].

Answer 1

Hint: Given expression has three fractions in trigonometric function form. To prove this identity we will start with LHS and conclude on RHS. For that we will cross multiply the terms on LHS. And then express \[{\sin ^2}\theta \] function in cos form. Further taking cos function common we will coms with a fraction that is equals to cot function. Taking reciprocal of which is a tangent function that is our RHS.

Complete step by step answer:
Given the identity to be proved is, \[\dfrac{{\sin \theta }}{{1 - \cos \theta }} - \dfrac{1}{{\sin \theta }} = \dfrac{1}{{\tan \theta }}\]
Now only focus on LHS,
Cross multiplying we get,
\[ = \dfrac{{\sin \theta .\sin \theta - \left( {1 - \cos \theta } \right)}}{{\sin \theta \left( {1 - \cos \theta } \right)}}\]
Solving the numerator,
\[ = \dfrac{{{{\sin }^2}\theta - 1 + \cos \theta }}{{\sin \theta \left( {1 - \cos \theta } \right)}}\]
 \[{\sin ^2}\theta \] can be written in cos form as,
 \[ = \dfrac{{1 - {{\cos }^2}\theta - 1 + \cos \theta }}{{\sin \theta \left( {1 - \cos \theta } \right)}}\]
Cancelling 1,
\[ = \dfrac{{\cos \theta - {{\cos }^2}\theta }}{{\sin \theta \left( {1 - \cos \theta } \right)}}\]
Taking cos common from the numerator,
\[ = \dfrac{{\cos \theta \left( {1 - \cos \theta } \right)}}{{\sin \theta \left( {1 - \cos \theta } \right)}}\]
Cancelling the bracket,
\[ = \cot \theta \]
As we know cot function is the reciprocal of tan. Thus we can write it as,
\[ = \dfrac{1}{{\tan \theta }}\]
And this is, \[ = RHS\]
Hence proved.

Note:
Note here that the way we start to solve the problem is a way important. Our approach really matters. Generally students are confused in the cross multiply process. This is nothing but taking the LCM. Here the RHS can simply tell that LHS can be a cot function on solving. And this is the clue we used to proceed in our problem.