Question

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

Answer 1

Hint: To prove the expression given in the question, we need to have an idea about a few trigonometric relations, which are, $\cos 2x=1-2{{\sin }^{2}}x$ and $\sin 2x=2\sin x\cos x$. By using these relations and some basic algebraic properties, we will prove the expression given in the question.

Complete step-by-step answer:

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

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

We know that $\cos 2x=1-2{{\sin }^{2}}x$ and $\sin 2x=2\sin x\cos x$. So, by using these trigonometric relations, we will get the LHS as,

$LHS=\dfrac{1-\left( 1-2{{\sin }^{2}}x \right)+\sin x}{2\left( \sin x \right)\left( \cos x \right)+\cos x}$

On further simplification, we can write it as,

$LHS=\dfrac{1-1+2{{\sin }^{2}}x+\sin x}{2\left( \sin x \right)\left( \cos x \right)+\cos x}$

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

Now, we can take $\sin x$ common from the numerator and also we can take $\cos x$

common from the denominator, so we will get the LHS as,

$LHS=\dfrac{\left( \sin x \right)\left( 2\sin x+1 \right)}{\left( \cos x \right)\left( 2\sin x+1 \right)}$

Here, we can see that $\left( 2\sin x+1 \right)$ is common in both the numerator and the denominator, so they will get cancelled and we will get the LHS as,

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

We know that, $\dfrac{\sin x}{\cos x}=\tan x$, so we can write the LHS as,

$LHS=\tan x$

Thus, we will get that $LHS=RHS$.

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

Note: In this question, the students can also make use of other trigonometric identities like, $\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}$ and also $\sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x}$. But when the students use these formulas, they might get confused after a point. So it is advisable not to use these trigonometric identities and it is better to use the trigonometric identities like, $\cos 2x=1-2{{\sin }^{2}}x$ and $\sin 2x=2\sin x\cos x$ to solve this question.