If $\alpha=\cos ^{-1}\left(\frac{3}{5}\right), \beta=\tan ^{-1}\left(\frac{1}{3}\right)$, where $0<\alpha, \beta<\frac{\pi}{2}$, then $\alpha-\beta$ is equal to:

(A) $\sin ^{-1}\left(\frac{9}{5 \sqrt{10}}\right)$

(B) $\cos ^{-1}\left(\frac{9}{5 \sqrt{10}}\right)$

(C) $\tan ^{-1}\left(\frac{9}{5 \sqrt{10}}\right)$

(D) $\tan ^{-1}\left(\frac{9}{14}\right)$

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Inverse Trigonometric Functions

α = cos^{- 1} (\frac{3}{5}), β = tan^{- 1} (\frac{1}{3})

, where

0 < α, β < \frac{π}{2}

, then

α - β

is equal to:

Answer: (A)

From

cos α = \frac{3}{5}

in the first quadrant,

sin α = \frac{4}{5}

tan α = \frac{4}{3}

, while

tan β = \frac{1}{3}

. Then

tan (α - β) = \frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{3} \cdot \frac{1}{3}} = \frac{1}{1 + \frac{4}{9}} = \frac{9}{13}

. Building a right triangle with opposite

9

and adjacent

13

gives hypotenuse

81 + 169 = 250 = 510

, so

sin (α - β) = \frac{9}{5 10}

and

α - β = sin^{- 1} (\frac{9}{5 10})

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option