Question

Question !

Prove that 2tan^-1(X) = sin^-1(2x/1+x²) ​

Answer 1

Step-by-step explanation:

Solution is in this attachment

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Hope it's helpful

@Raj

Answer 2

Question:

$2 {tan}^{ - 1} x = sin ^{ - 1} \frac{2x}{1 + {x}^{2} }$

Answer:

Let ,

${tan}^{ - 1} x = y \\ \\ x = tan \: y$

Now put the value of x in

$\frac{2x}{1 + {x}^{2} }$

we get,

$= > \frac{2tany}{1 + {tan}^{ 2}y } \\ \\ = > \frac{2 \frac{siny}{cosy} }{1 + \frac{ {sin}^{2}y }{ {cos}^{2} y} } \\ \\ = > \frac{2siny \: cosy}{ {cos}^{2}y + {sin}^{2} y } \\ \\ = > \frac{ sin2y}{1} \\ \\ = > sin2y$

Thus,

$2y = {sin}^{ - 1} \frac{2x}{1 + {x}^{2} }$

$2 {tan}^{ - 1} x = {sin}^{ - 1} \frac{2x}{1 + {x}^{2} }$

Hence proved.