If sin -12 a 1-a 2-cos -11-a 21-a 2-tan -12 - 1-x 2- where a- x - 0-1- then- the value of x isa 0b a2c ad 2 a 1- a 2

Sin 12a1 a2+cos 11 a21+a2=tan 12x1 x2 #8658;2tan 1a+2tan 1a=2tan 1x #8658;4tan 1a=2tan 1x #8658;2tan 1a=tan 1x #8658;tan 12a1 a2=tan 1x #8658;x=2a1 a2 ...

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Byju's Answer

Standard XII

Mathematics

Property 7

If sin -12 a ...

Question

If $\sin^{- 1} (\frac{2 a}{1 - a^{2}}) + \cos^{- 1} (\frac{1 - a^{2}}{1 + a^{2}}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}}), where a, x \in (0, 1)$ , then, the value of x is

(a) 0
(b) $\frac{a}{2}$
(c) a
(d) $\frac{2 a}{1 - a^{2}}$

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Solution

$\sin^{- 1} (\frac{2 a}{1 - a^{2}}) + \cos^{- 1} (\frac{1 - a^{2}}{1 + a^{2}}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}}) \Rightarrow 2 \tan^{- 1} a + 2 \tan^{- 1} a = 2 \tan^{- 1} x \Rightarrow 4 \tan^{- 1} a = 2 \tan^{- 1} x \Rightarrow 2 \tan^{- 1} a = \tan^{- 1} x \Rightarrow \tan^{- 1} (\frac{2 a}{1 - a^{2}}) = \tan^{- 1} x \Rightarrow x = \frac{2 a}{1 - a^{2}}$
Hence, the correct answer is option(d).

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Similar questions

If \sin^{- 1} \frac{2 a}{1 + a^{2}} - \cos^{- 1} \frac{1 - b^{2}}{1 + b^{2}} = \tan^{- 1} \frac{2 x}{1 - x^{2}}, then prove that x = \frac{a - b}{1 + a b}

If $t a n^{- 1} \frac{1}{a - 1} = t a n^{- 1} \frac{1}{x} + t a n^{- 1} \frac{1}{a^{2} - x + 1}$ , then x is

Q. If

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

then

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}}

Q. If

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}}

, then prove that

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

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If sin-12a1-a2+cos-11-a21+a2=tan-12x1-x2, where a, x∈0, 1, then, the value of x is (a) 0 (b) a2 (c) a (d) 2a1-a2

sin-12a1-a2+cos-11-a21+a2=tan-12x1-x2⇒2tan-1a+2tan-1a=2tan-1x⇒4tan-1a=2tan-1x⇒2tan-1a=tan-1x⇒tan-12a1-a2=tan-1x⇒x=2a1-a2 Hence, the correct answer is option(d).

If $\sin^{- 1} (\frac{2 a}{1 - a^{2}}) + \cos^{- 1} (\frac{1 - a^{2}}{1 + a^{2}}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}}), where a, x \in (0, 1)$ , then, the value of x is

(a) 0
(b) $\frac{a}{2}$
(c) a
(d) $\frac{2 a}{1 - a^{2}}$