Question

How do you prove $\dfrac{{\sin x}}{{1 - \cos x}} + \dfrac{{1 - \cos x}}{{\sin x}} = 2\cos {\text{ec}}x$?

Answer 1

Start with the left hand side and take the LCM to convert two fractions into one fraction. Use formulas ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab,{\text{ }}{\sin ^2}x + {\cos ^2}x = 1$ and $\dfrac{1}{{\sin x}} = {\text{cosec}}x$ to simplify the resultant expression and bring it in the form of right hand side.

Complete step-by-step solution:
According to the question, we have been given a trigonometric identity. We have to show how we can prove its validity.
The given trigonometric identity is:
$\Rightarrow \dfrac{{\sin x}}{{1 - \cos x}} + \dfrac{{1 - \cos x}}{{\sin x}} = 2\cos {\text{ec}}x{\text{ }}.....{\text{(1)}}$
We’ll start with left hand side and simplify the expression to bring it in the form of right hand side. So we have:
$\Rightarrow {\text{LHS}} = \dfrac{{\sin x}}{{1 - \cos x}} + \dfrac{{1 - \cos x}}{{\sin x}}$
If we take the LCM in the above expression by cross multiplying denominators, we’ll get:
$\Rightarrow {\text{LHS}} = \dfrac{{{{\sin }^2}x + {{\left( {1 - \cos x} \right)}^2}}}{{\sin x\left( {1 - \cos x} \right)}}$
Now, applying the formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ in the numerator, we’ll get:
$\Rightarrow {\text{LHS}} = \dfrac{{{{\sin }^2}x + 1 + {{\cos }^2}x - 2\cos x}}{{\sin x\left( {1 - \cos x} \right)}} \\\ \Rightarrow {\text{LHS}} = \dfrac{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + 1 - 2\cos x}}{{\sin x\left( {1 - \cos x} \right)}} \\\$
We know that according to another trigonometric identity:
$\Rightarrow {\sin ^2}x + {\cos ^2}x = 1$
Applying this in our above expression, we’ll get:
$\Rightarrow {\text{LHS}} = \dfrac{{1 + 1 - 2\cos x}}{{\sin x\left( {1 - \cos x} \right)}}$
Simplifying it further, we have:
$\Rightarrow {\text{LHS}} = \dfrac{{2 - 2\cos x}}{{\sin x\left( {1 - \cos x} \right)}} \\\ \Rightarrow {\text{LHS}} = \dfrac{{2\left( {1 - \cos x} \right)}}{{\sin x\left( {1 - \cos x} \right)}} \\\$
As we can observe, $\left( {1 - \cos x} \right)$ gets cancelled out from both the numerator and denominator. From this we’ll get:
$\Rightarrow {\text{LHS}} = \dfrac{2}{{\sin x}}$
From another trigonometric formula, we have:
$\Rightarrow \dfrac{1}{{\sin x}} = {\text{cosec}}x$
Using this in our expression, we’ll get:
$\Rightarrow {\text{LHS}} = 2{\text{cosec}}x{\text{ }}.....{\text{(2)}}$
Comparing this result with equation (1), we have:
$\Rightarrow {\text{LHS}} = {\text{RHS}}$
Hence we have proved the given trigonometric identity.

Note: In the above problem, we can also start with the right hand side and simplify it to bring it in the form of the left hand side. Although this will be slightly difficult for the current problem, given that only ${\text{2cosec}}x$ is present in the left hand side but conceptually it would be correct. And we can also use this method in other complicated problems. So we can either start from the left hand side or from the right hand side.