Question

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

Answer 1

Hint: We will use the double angle trigonometric formulas to simplify any one side of the equation or both sides so that the result for both should come same and therefore, we can prove that L.H.S. of the equation is equal to the R.H.S. Here we will use pythagorean identity ${\cos ^2}x + {\sin ^2}x = 1$ and some algebraic identity $(a + b)(a - b) = {a^2} - {b^2}$, ${(a + b)^2} = {a^2} + 2ab + {b^2}$ to prove the above equation.

Complete step by step answer:
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,
$\Rightarrow L.H.S. =\dfrac{{1 - \dfrac{{\sin x}}{{\cos x}}}}{{1 + \dfrac{{\sin x}}{{\cos x}}}}$

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

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

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

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

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

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

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

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

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,
$\Rightarrow L.H.S. = \dfrac{{{{\cos }^2}x - 2\sin x\cos x + {{\sin }^2}x}}{{\cos 2x}}$

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

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

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

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

Note:
We have three pythagorean identities, one we have used in the above problem and the other two are $1+cot^2\theta=csc^2\theta$ and $tan^2\theta+1=sec^2\theta$. We use them in almost all the trigonometric problems.