Considering the principal values of inverse trigonometric functions, the value of the expression $\tan \left(2 \sin ^{-1}\left(\frac{2}{\sqrt{13}}\right)-2 \cos ^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)$ is equal to :

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Inverse Trigonometric Functions

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

tan (2 sin^{- 1} (\frac{2}{13}) - 2 cos^{- 1} (\frac{3}{10}))

is equal to :

Answer: (B)

Put

θ = sin^{- 1} \frac{2}{13}

and

ϕ = cos^{- 1} \frac{3}{10}

, both in

[0, \frac{π}{2})

. Then

sin θ = \frac{2}{13}, cos θ = \frac{3}{13}

, so

tan θ = \frac{2}{3}

; and

cos ϕ = \frac{3}{10}, sin ϕ = \frac{1}{10}

, so

tan ϕ = \frac{1}{3}

. Using

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

tan 2 θ = \frac{2 \cdot \frac{2}{3}}{1 - \frac{4}{9}} = \frac{12}{5}, tan 2 ϕ = \frac{2 \cdot \frac{1}{3}}{1 - \frac{1}{9}} = \frac{3}{4} .

Therefore

tan (2 θ - 2 ϕ) = \frac{tan 2 θ - tan 2 ϕ}{1 + tan 2 θ tan 2 ϕ} = \frac{\frac{12}{5} - \frac{3}{4}}{1 + \frac{12}{5} \cdot \frac{3}{4}} = \frac{\frac{33}{20}}{\frac{14}{5}} = \frac{33}{56} .

The value of the expression is

\frac{33}{56}

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option