Question

How to verify $\dfrac{{\cos 2x}}{{(1 + \sin 2x)}} = \tan \left( {\dfrac{\pi }{4} - x} \right)$ ?

Answer 1

To solve this we need to know certain trigonometric identities. We take the left hand side of the equation and then we show it to be equal to the right hand side of the equation. We know the cosine and sine double angle formula $\cos 2x = {\cos ^2}x - {\sin ^2}x$ and $\sin 2x = 2\sin x.\cos x$. We also have Pythagoras relation between sine and cosine function. That is ${\sin ^2}x + {\cos ^2}x = 1$. Using this we can solve this.

Complete step by step answer:
Given, $\dfrac{{\cos 2x}}{{(1 + \sin 2x)}} = \tan \left( {\dfrac{\pi }{{4 - x}}} \right)$
Here $LHS = \dfrac{{\cos 2x}}{{(1 + \sin 2x)}}$ and $RHS = \tan \left( {\dfrac{\pi }{{4 - x}}} \right)$.
Let’s take LHS, that is
$LHS = \dfrac{{\cos 2x}}{{(1 + \sin 2x)}}$
$\dfrac{{\cos 2x}}{{(1 + \sin 2x)}}$
We have the identity $\cos 2x = {\cos ^2}x - {\sin ^2}x$ and ${\sin ^2}x + {\cos ^2}x = 1$,
$\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{({{\sin }^2}x + {{\cos }^2}x + \sin 2x)}}$
Also we have $\sin 2x = 2\sin x.\cos x$
$\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{({{\sin }^2}x + {{\cos }^2}x + 2\sin x.\cos x)}}$

We use the algebraic identity ${a^2} - {b^2} = (a + b)(a - b)$ and ${(a + b)^2} = {a^2} + {b^2} + 2ab$.
$\dfrac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{{{(\sin x + \cos x)}^2}}}$
Cancelling we have,
$\dfrac{{\cos x - \sin x}}{{\sin x + \cos x}}$
Now take $\cos x$ common in both the numerator and the denominator we have,
$\dfrac{{\cos x\left( {1 - \dfrac{{\sin x}}{{\cos x}}} \right)}}{{\cos x\left( {1 + \dfrac{{\sin x}}{{\cos x}}} \right)}}$
We know tangent is a ratio of sine and cosine,
$\dfrac{{1 - \tan x}}{{1 + \tan x}}$
We know that $\tan \left( {\dfrac{\pi }{4}} \right) = 1$,
$\dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) - \tan x}}{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan x}}$
But we know $\tan (a - b) = \dfrac{{\tan a - \tan b}}{{1 + \tan a.\tan b}}$ then we have,
$\tan \left( {\dfrac{\pi }{4} - x} \right)$.

That is $LHS = RHS$. Hence proved.

Note: We can also solve this by taking Right hand side of the equation and show it that it is equal to the left hand side of the equation. We use the same identity that what we used in above. Remember a graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant.