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How do you prove the identity \[\dfrac{{cosec \theta - 1}}{{\cot \theta }} = \dfrac{{\cot \theta }}{{cosec \theta + 1}}\]?

Last updated date: 24th Jun 2024
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Hint:Start from the left hand side and multiply both numerator and denominator by \[cosec \theta + 1\]. Then use the trigonometric formula ${cosec ^2}\theta - {\cot ^2}\theta = 1$ to simplify it further and bring it in the form of the right hand side.

Complete step by step answer:According to the question, we have been given a trigonometric identity and we are asked to prove it.
The trigonometric identity is:
\[ \Rightarrow \dfrac{{cosec \theta - 1}}{{\cot \theta }} = \dfrac{{\cot \theta }}{{cosec \theta + 1}}{\text{ }}.....{\text{(1)}}\]
We’ll start with the left hand side and apply trigonometric formulas to convert it in the form of right hand side.
So we have:
\[ \Rightarrow {\text{LHS}} = \dfrac{{cosec \theta - 1}}{{\cot \theta }}\]
If we multiply both numerator and denominator by \[cosec \theta + 1\], we’ll get:
   \Rightarrow {\text{LHS}} = \dfrac{{cosec \theta - 1}}{{\cot \theta }} \times \dfrac{{cosec \theta + 1}}{{cosec \theta + 1}} \\
   \Rightarrow {\text{LHS}} = \dfrac{{\left( {cosec \theta - 1} \right)\left( {cosec \theta + 1} \right)}}{{\cot \theta \left( {cosec \theta + 1} \right)}} \\
We know an important algebraic formula as:
\[ \Rightarrow \left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\]
Applying this formula in our trigonometric expression, we’ll get:
\[ \Rightarrow {\text{LHS}} = \dfrac{{{{cosec }^2}\theta - 1}}{{\cot \theta \left( {cosec \theta + 1} \right)}}\]
Now, we have a trigonometric formula as given below:
$ \Rightarrow {cosec ^2}\theta - {\cot ^2}\theta = 1$
Separating terms in this formula, we have:
$ \Rightarrow {cosec ^2}\theta - 1 = {\cot ^2}\theta $
So putting this value of ${cosec ^2}\theta - 1$ in our above trigonometric expression, we’ll get:
\[ \Rightarrow {\text{LHS}} = \dfrac{{{{\cot }^2}\theta }}{{\cot \theta \left( {cosec \theta + 1} \right)}}\]
On further simplification, \[\cot \theta \] will get cancel out from both numerator and denominator. Doing this we will get:
\[ \Rightarrow {\text{LHS}} = \dfrac{{\cot \theta }}{{cosec \theta + 1}}\]
Now comparing this final expression with the expression in equation (1), we can conclude that:
\[ \Rightarrow {\text{LHS}} = {\text{RHS}}\]
Therefore this is how we prove this identity.

In this problem, we can also start from the right hand side and simplify it to bring it in the form of the left hand side. The solution will still be correct conceptually.
Some of the widely used trigonometric formulas are:
(1) ${\sin ^2}\theta + {\cos ^2}\theta = 1$
(2) ${\sec ^2}\theta - {\tan ^2}\theta = 1$
(3) ${cosec ^2}\theta - {\cot ^2}\theta = 1$
The last formula we have already used in the above problem.