Question

How do you simplify $\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$

Answer 1

Hint: In this question we will start the proof from the left-hand side of the expression and use trigonometric identities such as the value of $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$ and that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and substitute them in the expression and simplify it to get the right-hand side of the expression hence proving that the expression is true.

Complete step-by-step answer:
We have the expression as:
$\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$
On taking the left-hand side of the expression, we get:
$\Rightarrow \cos 2x$
Now we know that $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$ on substituting it in the term, we get:
$\Rightarrow {{\cos }^{2}}x-{{\sin }^{2}}x$
Now this term can be expressed in terms of a fraction with the denominator as $1$ therefore, on rewriting the term, we get:
$\Rightarrow \dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{1}$
Now we know the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ therefore, on substituting this in the denominator of the expression, we get:
$\Rightarrow \dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\sin }^{2}}x+{{\cos }^{2}}x}$
On dividing both the numerator and denominator by ${{\cos }^{2}}x$,we get:
$\Rightarrow \dfrac{\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\cos }^{2}}x}}{\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\cos }^{2}}x}}$
Now on splitting the denominators in both the terms which are present in the numerator and denominator, we get:
$\Rightarrow \dfrac{\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x}-\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}{\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x}+\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}$
On cancelling the terms, we get:
$\Rightarrow \dfrac{1-\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}{1+\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}$
Now we know the trigonometric identity $\tan x=\dfrac{\sin x}{\cos x}$ therefore, on squaring both the sides of the expression, we get ${{\tan }^{2}}x=\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}$.
On substituting this in the above expression we get:
$\Rightarrow \dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$, which is the right-hand side of the expression, hence proved.

Note: Various trigonometric identities should be remembered while doing these types of sums. It is to be remembered that while doing any form of trigonometric proofs all the identities should be converted into the simpler form by converting them into $\sin $ and $\cos $, and then simplifying them further. The expressions which have two times of the angle are called double angles and various double angle formulas should be remembered.