The total number of real solutions of the equation \[ \theta=\tan ^{-1}(2 \tan \theta)-\frac{1}{2} \sin ^{-1}\left(\frac{6 \tan \theta}{9+\tan ^{2} \theta}\right) \] is (Here, the inverse trigonometric functions \(\sin ^{-1} x\) and \(\tan ^{-1} x\) assume values in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), respectively.)

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Inverse Trigonometric Functions

The total number of real solutions of the equation

θ = tan^{- 1} (2 tan θ) - \frac{1}{2} sin^{- 1} (\frac{6 tan θ}{9 + tan ^{2} θ})

is
(Here, the inverse trigonometric functions

sin^{- 1} x

and

tan^{- 1} x

assume values in

[- \frac{π}{2}, \frac{π}{2}]

and

(- \frac{π}{2}, \frac{π}{2})

, respectively.)

Answer: (C)

Let

α = \frac{1}{2} sin^{- 1} (\frac{6 t a n θ}{9 + t a n ^{2} θ})

.

Then

θ + α = tan^{- 1} (2 tan θ)

, so

tan (θ + α) = 2 tan θ

.

Expanding:

\frac{t a n θ + t a n α}{1 - t a n α t a n θ} = 2 tan θ

.

From

sin 2 α = \frac{6 t a n θ}{9 + t a n ^{2} θ} = \frac{2 t a n α}{1 + t a n ^{2} α}

, we get

3 tan θ (1 + tan^{2} α) = tan α (9 + tan^{2} θ)

.

Simplifying:

3 (tan θ - 3 tan α) = tan α tan θ (tan θ - 3 tan α)

.

This gives

tan θ = 3 tan α

tan θ = \frac{3}{t a n α}

.

Case I:

tan θ = 3 tan α

. Substituting back gives

tan θ = 0, 1, - 1

, yielding

θ = 0, \frac{π}{4}, - \frac{π}{4}

.

Case II:

tan θ = \frac{3}{t a n α}

. This leads to

tan^{2} θ = - \frac{3}{5}

, which has no real solutions.

Total: 3 solutions.

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Single correct option