Trigonometry · JEE Advanced previous year questions

2025 2 questions

1. Q32025 · 18 May 2025 · Paper 2

   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.)

   1. A.1
   2. B.2
   3. C.3
   4. D.5
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   Answer: C

2. Q152025 · 18 May 2025 · Paper 2· Numerical answer

   Let $$\alpha =\frac{1}{\sin 60^{\circ }\sin 61^{\circ }}+\frac{1}{\sin 62^{\circ }\sin 63^{\circ }}+\cdots +\frac{1}{\sin 118^{\circ }\sin 119^{\circ }}.$$ Then the value of $${\left(\frac{\mathrm{cosec}{1}^{\circ }}{\alpha }\right)}^2$$ is ____.

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   Answer: 3