Question

Prove the following: $\dfrac{\sin 2x}{1+\cos 2x}=\tan x$.

Answer 1

Hint: To prove the given expression, we need to know a few formulas like, $\sin 2x=2\sin x\cos x$ and $\cos 2x=2{{\cos }^{2}}x-1$ or we can also use other formulas like, $\sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x}$ and $\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$. So, by using any one pair of formulas from the above two pairs of formulas, we can prove the expression given in the question.

Complete step-by-step answer:

In this question, we have to prove that $\dfrac{\sin 2x}{1+\cos 2x}=\tan x$. In order to prove this expression, we will first consider the left hand side (LHS) of the expression. So, we can write it as,

$LHS=\dfrac{\sin 2x}{1+\cos 2x}$

We know that, $\sin 2x=2\sin x\cos x\ldots \ldots \ldots (i)$.

Also, we know that, $\cos 2x=2{{\cos }^{2}}x-1\ldots \ldots \ldots (ii)$.

Now, on substituting the values of $\sin 2x$, from equation (i) in the numerator of the LHS and also on substituting the values of $\cos 2x$ from equation (ii), in the denominator of the LHS, we will get,

$LHS=\dfrac{2\sin x\cos x}{1+2{{\cos }^{2}}x-1}$

On cancelling the similar terms from the denominator, we will get,

$LHS=\dfrac{2\sin x\cos x}{2{{\cos }^{2}}x}$

Now, we can cancel $2\cos x$ from the numerator and the denominator as they are common. So, we get the LHS as,

$LHS=\dfrac{\sin x}{\cos x}$

We know that, $\dfrac{\sin x}{\cos x}=\tan x$ therefore, we get the LHS as follows,

$LHS=\tan x$

We know that $\tan x$ is the RHS of the expression given in the question, therefore we get,

$LHS=RHS$

Hence, we have proved the expression that is $\dfrac{\sin 2x}{1+\cos 2x}=\tan x$.

Note: Another alternative method of solving this question is:

We know that the $LHS=\dfrac{\sin 2x}{1+\cos 2x}$. Also, we know that $\sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x}$ and $\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$. So, we get the LHS as, $LHS=\dfrac{\dfrac{2\tan x}{1+{{\tan }^{2}}x}}{1+\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}}$.

We will take the LCM of the denominator. So, we get the LHS as,

$LHS=\dfrac{\dfrac{2\tan x}{1+{{\tan }^{2}}x}}{\dfrac{1+{{\tan }^{2}}x+1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}}$

It can be also written as, $LHS=\dfrac{2\tan x\left( 1+{{\tan }^{2}}x \right)}{\left( 1+{{\tan }^{2}}x+1-{{\tan }^{2}}x \right)\left( 1+{{\tan }^{2}}x \right)}$.

Now, we can see that, $\left( 1+{{\tan }^{2}}x \right)$ is common in both numerator and denominator. So, we get,

$LHS=\dfrac{2\tan x}{1+{{\tan }^{2}}x+1-{{\tan }^{2}}x}$

Now, on cancelling the same terms, we will get,

$LHS=\dfrac{2\tan x}{2}$

This can be further simplified as,

$LHS=\tan x$

Therefore, we get that $LHS=RHS$.

Hence we have proved that, $\dfrac{\sin 2x}{1+\cos 2x}=\tan x$.