If sin-12a-1 - a2 - sin-12b-1 - b2 - 2 tan-1 x- then x -

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

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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 $x =$

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

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

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

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

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Solution

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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

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

x

[a, b ϵ (0, 1)]

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

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}

x = 1 + a + a^{2} + . . . . \infty, y = 1 + b + b^{2} + . . . . \infty, | a b | < 1,

1 + a b + a^{2} b^{2} + . . . . \infty =

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