Question

How do you use the double angle formula to verify: \[{\tan ^2}x = \dfrac{{1 - \cos 2x}}{{1 + \cos 2x}}\] ?

Answer 1

Hint: The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $\cos 2x = 2{\cos ^2}x - 1 = 1 - 2{\sin ^2}x = {\cos ^2}x - {\sin ^2}x$. Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem and proving the result given to us.

Complete step by step answer:
In the given problem, we have to prove a trigonometric equality that can be further used in many questions and problems as a direct result and has wide ranging applications. For proving the desired result, we need to have a good grip over the basic trigonometric formulae and identities.
Now, we need to make the left and right sides of the equation equal.
R.H.S. $ = \dfrac{{1 - \cos 2x}}{{1 + \cos 2x}}$
Now, we have to apply the double angle formula of cosine so as to simplify the numerator and denominator of the trigonometric rational function. Now, we know that $\cos 2x$ is equal to $\left( {{{\cos }^2}x - {{\sin }^2}x} \right)$.
We also know the trigonometric identity ${\cos ^2}x + {\sin ^2}x = 1$. So, we can have variations of the double angle formula of cosine as $\cos 2x = \left( {2{{\cos }^2}x - 1} \right) = \left( {1 - 2{{\sin }^2}x} \right) = \left( {{{\cos }^2}x - {{\sin }^2}x} \right)$.
We can use any variation as per our requirement as each formula has its own use and simplification method.
So, for simplifying the numerator of $\dfrac{{1 - \cos 2x}}{{1 + \cos 2x}}$, we can use the formula $\cos 2x = \left( {1 - 2{{\sin }^2}x} \right)$ and for the numerator of the rational trigonometric function, we can use the formula $\cos 2x = \left( {2{{\cos }^2}x - 1} \right)$. So, we get,
$ \Rightarrow \dfrac{{1 - \left( {1 - 2{{\sin }^2}x} \right)}}{{1 + \left( {2{{\cos }^2}x - 1} \right)}}$
Opening the brackets and simplifying the expression, we get,
$ \Rightarrow \dfrac{{1 - \left( {1 - 2{{\sin }^2}x} \right)}}{{1 + \left( {2{{\cos }^2}x - 1} \right)}}$
$ \Rightarrow \dfrac{{2{{\sin }^2}x}}{{2{{\cos }^2}x}}$
Cancelling the $2$ in numerator and denominator, we get,
$ \Rightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}$
Also, we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$. So, we get,
$ \Rightarrow {\tan ^2}x$
Now, L.H.S$ = {\tan ^2}x$
As the left side of the equation is equal to the right side of the equation, we have,
\[{\tan ^2}x = \dfrac{{1 - \cos 2x}}{{1 + \cos 2x}}\]
Hence, Proved.

Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations.