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JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Inverse Trigonometric Functions

tan (2 tan^{- 1} \frac{1}{5} + sec^{- 1} \frac{5}{2} + 2 tan^{- 1} \frac{1}{8})

is equal to:

(A) $1$
(B) $2$ ✓ Correct
(C) $\frac{1}{4}$
(D) $\frac{5}{4}$

Answer: (B)

Solution

First,

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

(since if

cos α = \frac{2}{5}

, then

tan α = \frac{1}{2}

). Using

tan^{- 1} a + tan^{- 1} b = tan^{- 1} \frac{a + b}{1 - ab}

:

tan^{- 1} \frac{1}{5} + tan^{- 1} \frac{1}{8} = tan^{- 1} \frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{40}} = tan^{- 1} \frac{13/40}{39/40} = tan^{- 1} \frac{1}{3}

. Thus

2 tan^{- 1} \frac{1}{5} + 2 tan^{- 1} \frac{1}{8} = 2 tan^{- 1} \frac{1}{3}

. Using

2 tan^{- 1} a = tan^{- 1} \frac{2 a}{1 - a ^{2}}

:

2 tan^{- 1} \frac{1}{3} = tan^{- 1} \frac{2/3}{8/9} = tan^{- 1} \frac{3}{4}

. Finally,

tan^{- 1} \frac{3}{4} + tan^{- 1} \frac{1}{2} = tan^{- 1} \frac{\frac{3}{4} + \frac{1}{2}}{1 - \frac{3}{8}} = tan^{- 1} \frac{5/4}{5/8} = tan^{- 1} 2

. Thus the answer is

2

.

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option

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