The total number of real solutions of the equation is \theta=\tan ^{-1}(2 \tan \theta)-\frac{1}{2} \sin ^{-1} (\frac{6 \tan \theta}{9+\tan ^2 \theta}) is (Here, the inverse trigonometric functions \sin ^{-1} x and \tan ^{-1} x assume values in [-\frac{\pi}{2}, \frac{\pi}{2}] and (-\frac{\pi}{2}, \frac{\pi}{2}), respectively). [JEE Adv 2025 P2]

Inverse Trigonometric Functions

Section: Mathematics

Question : 1 of 14

Marks: +1, -0

The total number of real solutions of the equation is
θ=tan−1(2‌tan‌θ)−‌

sin‌−1(‌

6‌tan‌θ

9+tan2θ

) is
(Here, the inverse trigonometric functions sin‌−1x and tan−1x assume values in [−‌

,‌

] and (−‌

,‌

), respectively).

[JEE Adv 2025 P2]

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