If x - 1 then 2 tan -1 x -sin -12 x -1- x 2-A- 4 tan -1 xB- 0C- -2D- -

The correct option is D πIf x ≥ 1 then 2 tan^ 1x =π +tan^ 1(2x/1 x^2 ) sin^ 1(2x/1+x^2 )=tan^ 1(2x/x^2 1 ) ∴ sum = π ...

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Properties Derived from Trigonometric Identities

If x ≥ 1 then...

Question

If $x \geq 1 t h e n 2 t a n^{- 1} x + s i n^{- 1} (\frac{2 x}{1 + x^{2}})$ =

4 t a n^{- 1} x

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0

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\frac{π}{2}

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π

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Similar questions

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

4 \tan^{- 1} x

\frac{π}{2}

π

If f (x) = 2 {tan}^{- 1} x + {sin}^{- 1} (\frac{2 x}{1 + x^{2}}), x > 1,

f (5)

{cot}^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x}) =

f (x) = s i n^{- 1} x + 2 t a n^{- 1} x + x^{2} + 4 x + 1

{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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