Question

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

Answer 1

Hint: To solve this question, we should know a few trigonometric relations, like $\sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x}$ and $\cos x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$. By using these relations we can prove the given expression.

Complete step-by-step answer:

In this question, we are asked to prove that $\dfrac{1+\sin 2x-\cos 2x}{1+\sin 2x+\cos 2x}=\tan x$. To prove this, first we will consider the left hand side (LHS) of the expression, that is $\dfrac{1+\sin 2x-\cos 2x}{1+\sin 2x+\cos 2x}$. We know that $\sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x}$ and $\cos x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$. So, we get the LHS as:

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

Now, we will take LCM in the numerator and denominator, so we get,

$\dfrac{\dfrac{\left( 1+{{\tan }^{2}}x \right)+\left( 2\tan x \right)-\left( 1-{{\tan }^{2}}x \right)}{\left( 1+{{\tan }^{2}}x \right)}}{\dfrac{\left( 1+{{\tan }^{2}}x \right)+\left( 2\tan x \right)+\left( 1-{{\tan }^{2}}x \right)}{\left( 1+{{\tan }^{2}}x \right)}}$

We can further write it as,

$\dfrac{\left[ \left( 1+{{\tan }^{2}}x \right)+\left( 2\tan x \right)-\left( 1-{{\tan }^{2}}x \right) \right]\left( 1+{{\tan }^{2}}x \right)}{\left[ \left( 1+{{\tan }^{2}}x \right)+\left( 2\tan x \right)+\left( 1-{{\tan }^{2}}x \right) \right]\left( 1+{{\tan }^{2}}x \right)}$

As we can see, the numerator and the denominator have $\left( 1+{{\tan }^{2}}x \right)$ in common, so they will get cancelled out and we will get,

$\dfrac{\left[ \left( 1+{{\tan }^{2}}x \right)+\left( 2\tan x \right)-\left( 1-{{\tan }^{2}}x \right) \right]}{\left[ \left( 1+{{\tan }^{2}}x \right)+\left( 2\tan x \right)+\left( 1-{{\tan }^{2}}x \right) \right]}$

 Or we can write it as,

$\dfrac{1+{{\tan }^{2}}x+2\tan x-1+{{\tan }^{2}}x}{1+{{\tan }^{2}}x+2\tan x+1-{{\tan }^{2}}x}$
We know that, equal terms with opposite signs get cancelled, so we get,

$\dfrac{2{{\tan }^{2}}x+2\tan x}{2+2\tan x}$

We can see that in the numerator, $2\left( \tan x \right)$ is common and in the denominator, 2 is common, so we can write it as,

$\dfrac{2\left( \tan x \right)\left( \tan x+1 \right)}{2\left( 1+\tan x \right)}$

Here, we can see that in the numerator and in the denominator $2\left( 1+\tan x \right)$ is common, so we can cancel the like terms, therefore we get, $\tan x$, which is equal to the right hand side (RHS) of the given expression in the question.

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

Note: We can also solve this question using another formulas like, $\sin 2x=2\sin x\cos x,\cos 2x=1-2{{\sin }^{2}}x$ or $\cos 2x=2{{\cos }^{2}}x-1$. Generally, the students get confused while solving, if they miss any steps. Students must be careful while taking the signs. Also the students must make sure not to make any mistakes while writing the trigonometric equations and must know the basic trigonometric functions.