$50 \tan \left(3 \tan ^{-1}\left(\frac{1}{2}\right)+2 \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)+ 4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1}(2 \sqrt{2})\right)$ is equal to ____.

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Inverse Trigonometric Functions

50 tan (3 tan^{- 1} (\frac{1}{2}) + 2 cos^{- 1} (\frac{1}{5})) + 42 tan (\frac{1}{2} tan^{- 1} (22))

is equal to ____.

Answer: 29

Note that

cos^{- 1} (\frac{1}{5}) = tan^{- 1} (2)

since if

cos α = \frac{1}{5}

, then

tan α = 2

.

Also,

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

(complementary angles).

So:

3 tan^{- 1} (\frac{1}{2}) + 2 cos^{- 1} (\frac{1}{5}) = 3 tan^{- 1} (\frac{1}{2}) + 2 tan^{- 1} (2)

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

Therefore:

tan (3 tan^{- 1} (\frac{1}{2}) + 2 cos^{- 1} (\frac{1}{5})) = tan (tan^{- 1} (\frac{1}{2}) + π) = \frac{1}{2}

.

For the second term, using the half-angle formula:

tan (\frac{1}{2} tan^{- 1} (22)) = \frac{1}{2}

.

Thus:

50 \cdot \frac{1}{2} + 42 \cdot \frac{1}{2} = 25 + 4 = 29

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