Letx=tanθorθ=tan−1x =sin−1(1+tan2θ2tanθ)=sin−1(sin2θ)=2θ=2tan−1xDifferentiating w.r.t x, We get:(1+x22)

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^ { - 1 } ( { 2 x } { 1 + x ^ { 2 } } )

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

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Text solutionVerified

Let $x = t an θ or θ = t a n^{- 1} x = s i n^{- 1} (\frac{2 t an θ}{1 + t a n ^{2} θ}) = s i n^{- 1} (s in 2 θ) = 2 θ = 2 t a n^{- 1} x$

Differentiating w.r.t x, We get:

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

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