If 0- x- 1- then sintan -1 1- x 2 2x -cos -1 1- x 2 1- x 2 is equal to

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Definition of Functions

If 0≤ x≤ 1,...

Question

If $0 \leq x \leq 1$ , then $sin {{tan}^{- 1} \frac{1 - x^{2}}{2 x} + {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}}$ is equal to

1

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Solution

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Similar questions

- 1 \leq x \leq 0

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

0 \leq x < 1

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

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

{sin}^{- 1} (\frac{2 x}{1 + x^{2}})

{cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})

1,

0 < x < 1

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

- 1 \leq x \leq 1

{cos}^{- 1} (2 x - 1 - x^{2}) + {sin}^{- 1} (y) = 0

x^{2} + y^{2} + x y

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