If \(\mathrm{k}=\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{2}{3}\right)\right)+\tan \left(\frac{1}{2} \sin ^{-1}\left(\frac{2}{3}\right)\right)\) then the number of solutions of the equation \(\sin ^{-1}(\mathrm{k} \mathrm{x}-1)=\sin ^{-1} \mathrm{x}-\cos ^{-1} \mathrm{x}\) is ______.

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Inverse Trigonometric Functions

k = tan (\frac{π}{4} + \frac{1}{2} cos^{- 1} (\frac{2}{3})) + tan (\frac{1}{2} sin^{- 1} (\frac{2}{3}))

then the number of solutions of the equation

sin^{- 1} (kx - 1) = sin^{- 1} x - cos^{- 1} x

is ______.

Answer: 1

Since

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

, writing

θ = \frac{1}{2} sin^{- 1} \frac{2}{3}

gives

k = tan (\frac{π}{2} - θ) + tan θ = cot θ + tan θ = \frac{2}{s i n 2 θ} = \frac{2}{2/3} = 3

. The equation

sin^{- 1} (3 x - 1) = sin^{- 1} x - cos^{- 1} x

requires

0 \leq x \leq \frac{2}{3}

. Taking sine of both sides,

3 x - 1 = sin (sin^{- 1} x - cos^{- 1} x) = 2 x^{2} - 1

, so

2 x^{2} - 3 x = 0

and

x = 0

x = \frac{3}{2}

. Only

x = 0

lies in

[0, \frac{2}{3}]

and it satisfies the original equation, so the number of solutions is

1

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Integer answer