Question

if A=tan^-1x,then the value of sin2A

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The value of $\sin 2A=\dfrac{2x}{1+x^2}$.

Step-by-step explanation:

We have,

$A=\tan^{-1}x$

⇒$\tan A=x$         ......... (1)

To find, the value of $\sin 2A=$

We know that,

$\sin 2A=2\sin A\cos A$

⇒ $\sin 2A=\dfrac{2\sin A\cos A}{\sin^2 A+\cos^2 A}$

⇒ $\sin 2A=\dfrac{\dfrac{2\sin A\cos A}{\cos^2 A}}{\dfrac{\sin^2 A+\cos^2 A}{\cos^2 A}}$

⇒ $\sin 2A=\dfrac{2\tan A}{1+\tan^2 A}$    ...... (2)

From equation (1) and (2), we get

$\sin 2A=\dfrac{2x}{1+x^2 }$

∴ $\sin 2A=\dfrac{2x}{1+x^2 }$

Hence, the value of  $\sin 2A=\dfrac{2x}{1+x^2}$.