Sin-12a1-a2-sin-12b1-b2-2tan-1x- then find the value of x-

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Domain and Range of Basic Inverse Trigonometric Functions

sin-12a1+a2+s...

Question

${sin}^{- 1} \frac{2 a}{1 + a^{2}} + {sin}^{- 1} \frac{2 b}{1 + b^{2}} = 2 {tan}^{- 1} x$ , then find the value of $x$ .

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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

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

{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

{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}}

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

{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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