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Question: How do you prove \(\dfrac{\left( 2\tan x \right)}{\left( 1+{{\tan }^{2}}x \right)}=\sin 2x?\)...

How do you prove (2tan⁡x)(1+tan⁡2x)=sin⁡2x?\dfrac{\left( 2\tan x \right)}{\left( 1+{{\tan }^{2}}x \right)}=\sin 2x?

Explanation

Solution

We solve this question simply by using the basic concepts or basic formulae of trigonometry. We use the formula tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} and sec⁡2x=1+tan⁡2x{{\sec }^{2}}x=1+{{\tan }^{2}}x to expand the left-hand side term and simplify it. Then we use the formula sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x in order to obtain the final answer.

Complete step by step answer:
In order to solve this question, let us use basic trigonometric rules. The given equation is (2tan⁡x)(1+tan⁡2x)=sin⁡2x.\dfrac{\left( 2\tan x \right)}{\left( 1+{{\tan }^{2}}x \right)}=\sin 2x. We are required to prove that the left-hand side of this expression is equal to the right-hand side of the expression. We first expand tan⁡x.\tan x.
⇒(2tan⁡x)(1+tan⁡2x)=sin⁡2x\Rightarrow \dfrac{\left( 2\tan x \right)}{\left( 1+{{\tan }^{2}}x \right)}=\sin 2x
We know that tan can be written in terms of sin and cos as tan⁡x=sin⁡xcos⁡x.\tan x=\dfrac{\sin x}{\cos x}. Substituting for tan⁡x\tan x in the numerator of left-hand side of the above equation,
⇒(2sin⁡xcos⁡x)(1+tan⁡2x)=sin⁡2x\Rightarrow \dfrac{\left( 2\dfrac{\sin x}{\cos x} \right)}{\left( 1+{{\tan }^{2}}x \right)}=\sin 2x
We also know that sec⁡2x=1+tan⁡2x{{\sec }^{2}}x=1+{{\tan }^{2}}x and we substitute this for the denominator term in the left-hand side of the above equation.
⇒(2sin⁡xcos⁡x)sec⁡2x=sin⁡2x\Rightarrow \dfrac{\left( 2\dfrac{\sin x}{\cos x} \right)}{{{\sec }^{2}}x}=\sin 2x
We know the relation between sec and cos as sec⁡x=1cos⁡x.\sec x=\dfrac{1}{\cos x}. Using this in the above equation,
⇒(2sin⁡xcos⁡x)1cos⁡2x=sin⁡2x\Rightarrow \dfrac{\left( 2\dfrac{\sin x}{\cos x} \right)}{\dfrac{1}{{{\cos }^{2}}x}}=\sin 2x
Now, taking the denominator term to the numerator by taking reciprocal,
⇒2sin⁡xcos⁡x×cos⁡2x1=sin⁡2x\Rightarrow 2\dfrac{\sin x}{\cos x}\times \dfrac{{{\cos }^{2}}x}{1}=\sin 2x
Cancelling the cos terms from the numerator and denominator,
⇒2sin⁡xcos⁡x=sin⁡2x\Rightarrow 2\sin x\cos x=\sin 2x
We know the standard relation that sin⁡2x=2sin⁡xcos⁡x.\sin 2x=2\sin x\cos x. Using this for the left-hand side of the equation,
⇒sin⁡2x=sin⁡2x\Rightarrow \sin 2x=\sin 2x
Hence, we have proved that LHS is equal to RHS.

Note: We need to know the general trigonometric relations such as the reciprocals and standard formulae in order to solve such problems. Care must be taken not to substitute the tan term in the denominator in terms of sin and cos as this will make the problem more complicated and difficult to solve. Knowing the formula that sec⁡2x=1+tan⁡2x{{\sec }^{2}}x=1+{{\tan }^{2}}x is useful for simplifying it to a great extent and avoiding complications.