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

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

Explanation

Solution

Hint: To prove the expression given in the question, we need to have an idea about a few trigonometric relations, which are, cos⁡2x=1−2sin⁡2x\cos 2x=1-2{{\sin }^{2}}x and sin⁡2x=2sin⁡xcos⁡x\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 1−cos⁡2x+sin⁡xsin⁡2x+cos⁡x=tan⁡x\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=1−cos⁡2x+sin⁡xsin⁡2x+cos⁡xLHS=\dfrac{1-\cos 2x+\sin x}{\sin 2x+\cos x}

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

LHS=1−(1−2sin⁡2x)+sin⁡x2(sin⁡x)(cos⁡x)+cos⁡xLHS=\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=1−1+2sin⁡2x+sin⁡x2(sin⁡x)(cos⁡x)+cos⁡xLHS=\dfrac{1-1+2{{\sin }^{2}}x+\sin x}{2\left( \sin x \right)\left( \cos x \right)+\cos x}

LHS=2sin⁡2x+sin⁡x2sin⁡xcos⁡x+cos⁡xLHS=\dfrac{2{{\sin }^{2}}x+\sin x}{2\sin x\cos x+\cos x}

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

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

LHS=(sin⁡x)(2sin⁡x+1)(cos⁡x)(2sin⁡x+1)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 (2sin⁡x+1)\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=sin⁡xcos⁡xLHS=\dfrac{\sin x}{\cos x}

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

LHS=tan⁡xLHS=\tan x

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

Hence we have proved the expression given in the question, that is, 1−cos⁡2x+sin⁡xsin⁡2x+cos⁡x=tan⁡x\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=1−tan⁡2x1+tan⁡2x\cos 2x=\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} and also sin⁡2x=2tan⁡x1+tan⁡2x\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=1−2sin⁡2x\cos 2x=1-2{{\sin }^{2}}x and sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x to solve this question.