Question

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

Answer 1

Hint:Here we have to prove whether the right-hand side of the equation is equal to the left-hand side of the equation which can be done by simplifying any one side of the equality or both the sides using trigonometric identities and formulas.

Complete step by step solution:
Some basic trigonometric ratios are, $\tan x = \dfrac{{\sin x}}{{\cos x}}$, $\sin x = \dfrac{1}{{\cos ecx}}$ and $\cos x = \dfrac{1}{{\sec x}}$

We have to prove $\dfrac{{1 + \tan x}}{{1 - \tan x}} = \dfrac{{1 + \sin 2x}}{{\cos 2x}}$ where
L.H.S. = $\dfrac{{1 + \tan x}}{{1 - \tan x}}$ and
R.H.S. = $\dfrac{{1 + \sin 2x}}{{\cos 2x}}$

First, we will simplify the
L.H.S. = $\dfrac{{1 + \tan x}}{{1 - \tan x}}$,

From the ratio, $\tan x = \dfrac{{\sin x}}{{\cos x}}$ we substitute the value of $\tan x$ in the L.H.S. of the equation,
L.H.S. = $\dfrac{{1 + \dfrac{{\sin x}}{{\cos x}}}}{{1 - \dfrac{{\sin x}}{{\cos x}}}}$

Taking the lowest common multiple in numerator and denominator, i.e., multiplying and dividing the term $1$ by $\cos x$,
L.H.S. = $\dfrac{{\dfrac{{\cos x}}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{{\cos x}}{{\cos x}} - \dfrac{{\sin x}}{{\cos x}}}}$

Since the denominators is same, we can add and subtract the numerator part,
L.H.S. = $\dfrac{{\dfrac{{\cos x + \sin x}}{{\cos x}}}}{{\dfrac{{\cos x - \sin x}}{{\cos x}}}}$

Cancelling $\cos x$ from denominator of both numerator and denominator, we get,
L.H.S. = $\dfrac{{\cos x + \sin x}}{{\cos x - \sin x}}$

Multiplying and dividing by $\cos x + \sin x$,
L.H.S. = $\dfrac{{\cos x + \sin x}}{{\cos x - \sin x}} \times \dfrac{{\cos x + \sin x}}{{\cos x + \sin x}}$

Multiplying the numerator and denominator part,
L.H.S. = $\dfrac{{(\cos x + \sin x)(\cos x + \sin x)}}{{(\cos x - \sin x)(\cos x + \sin x)}}$

In the numerator both terms are same, so we can write them as $a.a = {a^2}$,
L.H.S. = $\dfrac{{{{(\cos x + \sin x)}^2}}}{{(\cos x - \sin x)(\cos x + \sin x)}}$

The denominator part is of the form $(a + b)(a - b)$ so we can substitute $(a + b)(a - b) = {a^2} - {b^2}$,
L.H.S. = $\dfrac{{{{(\cos x + \sin x)}^2}}}{{({{\cos }^2}x - {{\sin }^2}x)}}$

Expanding the numerator in the form ${(a + b)^2} = {a^2} + 2ab + {b^2}$,
L.H.S. = $\dfrac{{{{\cos }^2}x + 2\sin x\cos x + {{\sin }^2}x}}{{({{\cos }^2}x - {{\sin }^2}x)}}$

The term ${\cos ^2}x - {\sin ^2}x$ is a double angle formula of $\cos 2x$, so substituting ${\cos ^2}x - {\sin ^2}x = \cos 2x$, we get,
L.H.S. = $\dfrac{{{{\cos }^2}x + 2\sin x\cos x + {{\sin }^2}x}}{{\cos 2x}}$

We can write this equation as,
L.H.S. = $\dfrac{{{{\cos }^2}x + {{\sin }^2}x + 2\sin x\cos x}}{{\cos 2x}}$

We know, ${\cos ^2}x + {\sin ^2}x = 1$ which is a trigonometric identity, so substituting this value,
L.H.S. = $\dfrac{{1 + 2\sin x\cos x}}{{\cos 2x}}$

We know that $\sin 2x = 2\sin x\cos x$ which is double angle formula, therefore, substituting this value,
L.H.S. = $\dfrac{{1 + \sin 2x}}{{\cos 2x}}$ = R.H.S.

Therefore, we have proved L.H.S. = R.H.S.

Hence proved $\dfrac{{1 + \tan x}}{{1 - \tan x}} = \dfrac{{1 + \sin 2x}}{{\cos 2x}}$.

Note:
While simplifying trigonometric problems, one should have a proper knowledge of all the trigonometric formulas and identities and basic arithmetic formulas such as $(a + b)(a - b) = {a^2} - {b^2}$, ${(a + b)^2} = {a^2} + 2ab + {b^2}$ , etc.