If sin-12a-1-a2-sin-12b-1-b2-2tan-1x- then x is equal to -a-b -0-1-

The correct option is A a+b/1 ab sin^ 1(2a1+a^2)+sin^ 1(2b1+b^2) =tan^ 1(2a1 a^2)+tan^ 1(2b1 b^2) =2tan^ 1(a)+2tan^ 1(b) =2tan^ 1(a+b1 ...

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Definition of Limit

If sin-12a/...

Question

If $s i n^{- 1} (\frac{2 a}{1 + a^{2}}) + s i n^{- 1} (\frac{2 b}{1 + b^{2}}) = 2 t a n^{- 1} x$ , then $x$ is equal to $[a, b ϵ (0, 1)]$

\frac{a - b}{1 + a b}

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\frac{b}{1 + a b}

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\frac{b}{1 - a b}

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\frac{a + b}{1 - a b}

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s i n^{- 1} (\frac{2 a}{1 + a^{2}}) + s i n^{- 1} (\frac{2 b}{1 + b^{2}}) = 2 t a n^{- 1} x

{sin}^{- 1} (\frac{2 a}{1 + a^{2}}) + {sin}^{- 1} (\frac{2 b}{1 + b^{2}}) = 2 {tan}^{- 1} x

s i n^{- 1} \frac{2 a}{1 + a^{2}} - c o s^{- 1} \frac{1 - b^{2}}{1 + b^{2}} = {t a n}^{- 1} \frac{2 x}{1 - x^{2}}

x = \frac{a - b}{1 + a b}

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

x

a b < 1

{cos}^{- 1} (\frac{1 - a^{2}}{1 + a^{2}}) + {cos}^{- 1} (\frac{1 - b^{2}}{1 + b^{2}}) = 2 {tan}^{- 1} x

x

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