The total number of real solutions of the equation = ^ -1 (2 )- 1 2 ^ -1 ( 6 9+ ^ 2 ) is…
Mathematics · JEE Advanced · NTA Exams — Trigonometry
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−1x and tan−1x assume values in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),\) respectively.).
\(\theta=\tan ^{-1}(2 \tan \theta)-\frac{1}{2} \sin ^{-1}\left(\frac{6 \tan \theta}{9+\tan ^{2} \theta}\right)\) is
(Here, the inverse trigonometric functions sin−1x and tan−1x assume values in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),\) respectively.).
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Answer
(C) 3
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