Question

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

Answer 1

In order to prove the given equation, we will consider any side of the given equation and simplify it using the different trigonometric identities and formulae. Through the successful application of identities and formulae, we simplify the expression until we get the same expression as on the other side of the equation. Here the problems revolve around trigonometric identities.

Complete step-by-step solution:
We need to prove: $\dfrac{{1 - \cos x}}{{\sin x}} = \dfrac{{\sin x}}{{1 + \cos x}}$
Proof:
We consider the LHS first: $\dfrac{{1 - \cos x}}{{\sin x}}$
Now, we express the angle ‘x’ in terms of $\left( {\dfrac{x}{2}} \right)$:
$= \dfrac{{1 - \cos \left( {2 \times \dfrac{x}{2}} \right)}}{{\sin \left( {2 \times \dfrac{x}{2}} \right)}}$ …………………….(1)
Again, we have the trigonometric identities:
${\sin ^2}\left( \theta \right) + {\cos ^2}\left( \theta \right) = 1$ …………………….(2)
$\cos \left( {2\theta } \right) = {\cos ^2}\left( \theta \right) - {\sin ^2}\left( \theta \right)$ ……………………(3)
$\sin \left( {2\theta } \right) = 2\sin \left( \theta \right)\cos \left( \theta \right)$ …………………….(4)
Now, replacing the ‘$\theta$’ in equations (2), (3) and (4) with $\left( {\dfrac{x}{2}} \right)$, we get:
${\sin ^2}\left( {\dfrac{x}{2}} \right) + {\cos ^2}\left( {\dfrac{x}{2}} \right) = 1$ …………………….(5)
$\cos \left( {2 \times \dfrac{x}{2}} \right) = {\cos ^2}\left( {\dfrac{x}{2}} \right) - {\sin ^2}\left( {\dfrac{x}{2}} \right)$ ……………………(6)
$\sin \left( {2 \times \dfrac{x}{2}} \right) = 2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)$ …………………….(7)
Thus, substituting the corresponding values from equations (5), (6) and (7) in expression (1), we get:
$= \dfrac{{{{\sin }^2}\left( {\dfrac{x}{2}} \right) + {{\cos }^2}\left( {\dfrac{x}{2}} \right) - \left[ {{{\cos }^2}\left( {\dfrac{x}{2}} \right) - {{\sin }^2}\left( {\dfrac{x}{2}} \right)} \right]}}{{2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)}}$
Simplifying further, we have:

$$= \dfrac{{{{\sin }^2}\left( {\dfrac{x}{2}} \right) + {{\cos }^2}\left( {\dfrac{x}{2}} \right) - {{\cos }^2}\left( {\dfrac{x}{2}} \right) + {{\sin }^2}\left( {\dfrac{x}{2}} \right)}}{{2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)}} \\\ = \dfrac{{2{{\sin }^2}\left( {\dfrac{x}{2}} \right)}}{{2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)}} \\\$$

Now, splitting the product in the numerator, we have:
$= \dfrac{{2 \times \sin \left( {\dfrac{x}{2}} \right) \times \sin \left( {\dfrac{x}{2}} \right)}}{{2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)}}$
We now cancel the common $\sin \left( {\dfrac{x}{2}} \right)$ term from the numerator as well as the denominator. Additionally, we multiply $\cos \left( {\dfrac{x}{2}} \right)$ to both the numerator and denominator to get:
$= \dfrac{{2 \times \sin \dfrac{x}{2} \times \cos \dfrac{x}{2}}}{{2 \times \cos \dfrac{x}{2} \times \cos \dfrac{x}{2}}}$ ……………………(8)
Now, using the formula in equation (7) in expression (8), and also simplifying the denominator, we get:
$= \dfrac{{\sin \left( {2 \times \dfrac{x}{2}} \right)}}{{2{{\cos }^2}\left( {\dfrac{x}{2}} \right)}}$
Simplifying further:
$= \dfrac{{\sin x}}{{{{\cos }^2}\left( {\dfrac{x}{2}} \right) + {{\cos }^2}\left( {\dfrac{x}{2}} \right)}}$
$= \dfrac{{\sin x}}{{{{\cos }^2}\left( {\dfrac{x}{2}} \right) + {{\sin }^2}\left( {\dfrac{x}{2}} \right) + {{\cos }^2}\left( {\dfrac{x}{2}} \right) - {{\sin }^2}\left( {\dfrac{x}{2}} \right)}}$ ………………………(9)
Using equations (5) and (6) in the above expression (9), we now have:
$= \dfrac{{\sin x}}{{1 + \cos \left( {2 \times \dfrac{x}{2}} \right)}}$
$= \dfrac{{\sin x}}{{1 + \cos x}}$, which is nothing but the expression on the other side of the expression, that is, on the right side of the expression.
Hence, proved.

Note: The alternative way of solving the same problem would be to take a common denominator with $\tan x$ and $\sec x$ terms, without converting them into trigonometric ratios containing the $\sin x$ and $\cos x$ terms. Then, we will make use of the trigonometric identity $1 + {\tan ^2}x = {\sec ^2}x$. Further simplification is as usual.

To prove a trigonometric equation, we should tend to start with the more complicated side of the equation, and keep simplifying it until it is transformed into the same expression as on the other side of the given equation. In some cases, we can also try to simplify both sides of the equation and arrive at a common expression to prove their equality. The various procedures of solving a trigonometric equation are: expanding the expressions, making use of the identities, factoring the expressions or simply using basic algebraic strategies to obtain the desired results.