If cos-1 1-x21-x2 -sin-1 2x1-x2 -K- x - - -1-0 - then the value of K is

The correct option is A 0 Substituting x=tanA Hence cos^ 1(1 tan^2A1+tan^2A)+sin^ 1(2tanA1+tan^2A) =cos^ 1(cos2A)+sin^ 1(sin2A) Now, xϵ[ ...

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Sin(A+B)Sin(A-B)

If cos-1 1-...

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If $c o s^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + s i n^{- 1} (\frac{2 x}{1 + x^{2}}) = K, \forall x ϵ [- 1, 0],$ then the value of K is

\frac{π}{2}

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- \frac{π}{2}

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3 s i n^{- 1} \frac{2 x}{1 + x^{2}} - 4 c o s^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 t a n^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{3}

3 s i n^{- 1} \frac{2 x}{1 + x^{2}} - 4 c o s^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 t a n^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{3}

s i n^{- 1} (x - 1) + c o s^{- 1} (x - 3) + t a n^{- 1} (\frac{x}{2 - x^{2}}) = c o s^{- 1} k + π

k

3 {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) - 4 {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + 2 {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3}

x

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

s i n [(1 / 5) {c o s}^{- 1} x] = 1

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

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