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

Explanation for the correct option.Given that, sin^ 1[2a/(1+a^2)]+sin^ 1[2b/(1+b^2)]=2tan^ 1(x)We know the identity, 2tan^ 1a=sin^ 12a/1+a^2.So, we have[ ...

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Standard XII

Mathematics

Properties of Determinants

If sin-1[2a/1...

Question

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

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

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

Explanation for the correct option.
Given that, $\sin^{- 1} [\frac{2 a}{(1 + a^{2})}] + \sin^{- 1} [\frac{2 b}{(1 + b^{2})}] = 2 \tan^{- 1} (x)$
We know the identity, $2 \tan^{- 1} a = \sin^{- 1} \frac{2 a}{1 + a^{2}}$ .
So, we have
$\begin{array}{rcl} 2 \tan^{- 1} a + 2 \tan^{- 1} b & = & 2 \tan^{- 1} x \\ \Rightarrow & \tan^{- 1} a + \tan^{- 1} b = \tan^{- 1} x \\ \Rightarrow & \tan^{- 1} \frac{a + b}{1 - a b} = \tan^{- 1} x \\ \Rightarrow & x = \frac{a + b}{1 - a b} \end{array}$
Hence, option D is correct.

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If sin-12a1+a2+sin-12b1+b2=2tan-1(x), then x=

The correct option is D a+b1-abExplanation for the correct option.Given that, sin-12a1+a2+sin-12b1+b2=2tan-1(x)We know the identity, 2tan-1a=sin-12a1+a2.So, we have2tan-1a+2tan-1b=2tan-1x⇒tan-1a+tan-1b=tan-1x⇒tan-1a+b1-ab=tan-1x⇒x=a+b1-abHence, option D is correct.

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