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Question: If \(A = {\tan ^{ - 1}}(x)\) then the value of \(\sin 2A = \) ? A) \(\dfrac{{2x}}{{\sqrt {1 - {x^2...

If A=tan⁡−1(x)A = {\tan ^{ - 1}}(x) then the value of sin⁡2A=\sin 2A = ?
A) 2x1−x2\dfrac{{2x}}{{\sqrt {1 - {x^2}} }}
B) 2a1−x2\dfrac{{2a}}{{1 - {x^2}}}
C) 2x1+x2\dfrac{{2x}}{{1 + {x^2}}}
D) None of these

Explanation

Solution

In order to find a solution for this question, you have to know about basic formulas of trigonometry and also you have to know how to draw trigonometric triangles. After that, rewrite the given equation in terms of xx. Then write a trigonometric formula for sin⁡2A\sin 2A. After doing that, divide that sin⁡2A\sin 2A formula by cos⁡2A{\cos ^2}A and put the value of xx and you will see your answer.

Complete step by step answer:
First of all, rewrite given equation in terms of xx ,
Given equation is
⇒A=tan⁡−1(x)\Rightarrow A = {\tan ^{ - 1}}(x)
Convert it into in terms of xx ,
⇒x=tan⁡A\Rightarrow x = \tan A
Now, let’s see what the special formula is for sin⁡2A\sin 2A ,
⇒sin⁡2A=2sin⁡Acos⁡A\Rightarrow \sin 2A = 2\sin A\cos A
We can write above formula as also like bellow one,
⇒sin⁡2A=2sin⁡Acos⁡A1\Rightarrow \sin 2A = \dfrac{{2\sin A\cos A}}{1}
You know that 1=sin⁡2A+cos⁡2A1 = {\sin ^2}A + {\cos ^2}A
So we can write,
⇒sin⁡2A=2sin⁡Acos⁡Asin⁡2A+cos⁡2A\Rightarrow \sin 2A = \dfrac{{2\sin A\cos A}}{{{{\sin }^2}A + {{\cos }^2}A}}
Now, divide above equation with respect to cos⁡2A{\cos ^2}A and we will get,
⇒sin⁡2A=2sin⁡Acos⁡Acos⁡2Asin⁡2A+cos⁡2Acos⁡2A\Rightarrow \sin 2A = \dfrac{{\dfrac{{2\sin A\cos A}}{{{{\cos }^2}A}}}}{{\dfrac{{{{\sin }^2}A + {{\cos }^2}A}}{{{{\cos }^2}A}}}}
From further simplification we will get,
⇒sin⁡2A=2tan⁡A1+tan⁡2A\Rightarrow \sin 2A = \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}
Now, as you see x=tan⁡Ax = \tan A
So, put value of xx in above equation and we will get,
⇒sin⁡2A=2x1+x2\Rightarrow \sin 2A = \dfrac{{2x}}{{1 + {x^2}}}
See, this equation is also there in our options.
Therefore, the correct answer to the given problem is option (A).

Note:
If you think the above method is a little bit complicated for this kind of small problems then you can use the method of trigonometric triangle. In that method you have to draw a triangle for x=tan⁡Ax = \tan A and after that you have to find the value for sin⁡A\sin A and cos⁡A\cos A . after finding that use the special formula for sin⁡2A\sin 2A as we see it in above method and you will get same answer as above method.