If sin -1 2a 1- a 2 -sin -1 2b 1-a 2 -2tan -1 x- then x is equal to

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Substitution Method to Remove Indeterminate Form

If sin -1 ...

Question

If ${sin}^{- 1} (\frac{2 a}{1 + a^{2}}) + {sin}^{- 1} (\frac{2 b}{1 + a^{2}}) = 2 {tan}^{- 1} x$ , then $x$ is equal to

\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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Solution

The correct option is D $\frac{a + b}{1 - a b}$
Let $a = t a n A$ and $b = t a n B$
Then the given expression reduces to
$s i n^{- 1} (s i n 2 A) + s i n^{- 1} (s i n 2 B) = 2 t a n^{- 1} (x)$
$2 (A + B) = 2 t a n^{- 1} (x)$
$t a n^{- 1} (x) = A + B$
$t a n^{- 1} (x) = t a n^{- 1} (a) + t a n^{- 1} (a)$
$t a n^{- 1} (x) = t a n^{- 1} (\frac{a + b}{1 - a b})$
$x = \frac{a + b}{1 - a b}$

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If sin−1(2a1+a2)+sin−1(2b1+a2)=2tan−1x, then x is equal to

The correct option is D a+b1−abLet a=tanA and b=tanBThen the given expression reduces tosin−1(sin2A)+sin−1(sin2B)=2tan−1(x)2(A+B)=2tan−1(x)tan−1(x)=A+Btan−1(x)=tan−1(a)+tan−1(a)tan−1(x)=tan−1(a+b1−ab)x=a+b1−ab

If ${sin}^{- 1} (\frac{2 a}{1 + a^{2}}) + {sin}^{- 1} (\frac{2 b}{1 + a^{2}}) = 2 {tan}^{- 1} x$ , then $x$ is equal to