Let >x>=>tanθ>or>θ>=>tan^ 1x #160; => sin^ 1(2tanθ>1+tan^2θ>) =sin^ 1(sin2θ) =2θ =2tan^ 1x Differentiating ...

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Basic Inverse Trigonometric Functions

sin - 1 2 x...

Question

${sin}^{- 1} (\frac{2 x}{1 + x^{2}})$

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Solution

Let $x = t a n θ o r θ = t a n^{- 1} x = s i n^{- 1} (\frac{2 t a n θ}{1 + t a n^{2} θ}) = s i n^{- 1} (s i n 2 θ) = 2 θ = 2 t a n^{- 1} x$
Differentiating w.r.t x, We get:
$(\frac{2}{1 + x^{2}})$

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