Considering only the principal values of the inverse trigonometric functions, the value of $$\frac{3}{2} \cos ^{-1} \sqrt{\frac{2}{2+\pi^{2}}}+\frac{1}{4} \sin ^{-1} \frac{2 \sqrt{2} \pi}{2+\pi^{2}}+\tan ^{-1} \frac{\sqrt{2}}{\pi}$$ is ____.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Inverse Trigonometric Functions

Considering only the principal values of the inverse trigonometric functions, the value of

\frac{3}{2} cos^{- 1} \frac{2}{2 + π ^{2}} + \frac{1}{4} sin^{- 1} \frac{2 2 π}{2 + π ^{2}} + tan^{- 1} \frac{2}{π}

is ____.

Answer: 2.355

Let

t = \frac{π}{2}

. Then

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

.

For the second term, note that

\frac{2 2 π}{2 + π ^{2}} = \frac{2 t}{1 + t ^{2}}

where

t = \frac{π}{2} > 1

. Using the identity

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

for

x \geq 1

, we get

sin^{- 1} (\frac{2 2 π}{2 + π ^{2}}) = π - 2 tan^{- 1} t

.

For the third term,

tan^{- 1} \frac{2}{π} = cot^{- 1} \frac{π}{2} = cot^{- 1} t

.

Substituting:

\frac{3}{2} tan^{- 1} t + \frac{1}{4} (π - 2 tan^{- 1} t) + cot^{- 1} t = (\frac{3}{2} - \frac{1}{2}) tan^{- 1} t + \frac{π}{4} + cot^{- 1} t

= tan^{- 1} t + cot^{- 1} t + \frac{π}{4} = \frac{π}{2} + \frac{π}{4} = \frac{3 π}{4} \approx 2.356

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Numerical answer