If sin -12 a -1- a 2-sin -12 b -1- b 2-2 tan -1 x- then x is equal toA- a-1-1-a tB- D-1-a tC- b-1-a bD- a - b -1- ab

The correct option is D a+b/1 abPutting a=tan α, b=tan β, the given expression ⇒ sin^ 1sin2 α +sin^ 1sin 2 β =2 tan^ 1x ⇒ 2 α +2 β =2 tan^ 1x ⇒ tan^ 1a +tan^ 1 ...

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Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

If sin -12 a ...

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

\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 $\frac{a + b}{1 - a b}$
Putting $a = t a n α, b = t a n β$ , the given expression
$\Rightarrow s i n^{- 1} s i n 2 α + s i n^{- 1} s i n 2 β = 2 t a n^{- 1} x \Rightarrow 2 α + 2 β = 2 t a n^{- 1} x \Rightarrow t a n^{- 1} a + t a n^{- 1} b = t a n^{- 1} x \Rightarrow t a n^{- 1} \frac{a + b}{1 - a b} = t a n^{- 1} x \Rightarrow x = \frac{a + b}{1 - a b}$

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Similar questions

Q. If

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

, then x is equal to

Q. 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)]

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

, then find the value of

x

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

Inverse circular functions,Principal values of

{sin}^{- 1} x, {c o s}^{- 1} x, {tan}^{- 1} x

{tan}^{- 1} x + {tan}^{- 1} y = {tan}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {tan}^{- 1} \frac{x + y}{1 - x y}

x y > 1

(a) If

{sin}^{- 1} \frac{2 p}{1 + p^{2}} - {cos}^{- 1} \frac{1 - q^{2}}{1 + q^{2}} = {tan}^{- 1} \frac{2 x}{1 - x^{2}}

then prove that

x = \frac{p - q}{1 + p q}

(b) Solve for x

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

tan [\frac{1}{2} {sin}^{- 1} \frac{2 a}{1 + a^{2}} + \frac{1}{2} {cos}^{- 1} \frac{1 - a^{2}}{1 + a^{2}}] = \frac{2 a}{1 - a^{2}}

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If sin−1(2a1+a2)+sin−1(2b1+b2)=2tan−1x, then x is equal to

The correct option is D a+b1−abPutting a=tanα,b=tanβ, the given expression ⇒sin−1sin2α+sin−1sin2β=2tan−1x⇒2α+2β=2tan−1x⇒tan−1a+tan−1b=tan−1x⇒tan−1a+b1−ab=tan−1x⇒x=a+b1−ab

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