Aspire Faculty ID #18085 · Topic: JEE Main 2026 (24 January Evening Shift) · Just now
JEE Main 2026 (24 January Evening Shift)

The number of elements in the set $ { x \in [0,180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ)\tan x \tan(x - 50^\circ) } $ is ______

Solution

$ \frac{\tan(x + 100^\circ)}{\tan x} = \tan(x + 50^\circ)\tan(x - 50^\circ) $ $ \frac{\sin(x + 100^\circ)\cos x}{\cos(x + 100^\circ)\sin x} = \frac{\sin(x + 50^\circ)\sin(x - 50^\circ)}{\cos(x + 50^\circ)\cos(x - 50^\circ)} $ Apply $ C $ & $ D $ $ \frac{\sin(2x + 100^\circ)}{\sin 100^\circ} = \frac{\cos 100^\circ}{-\cos 2x} $ $ 2\sin(2x + 100^\circ)\cos 2x + \sin 100^\circ + \sin 200^\circ = 0 $ $ \sin(4x + 100^\circ) + \sin 100^\circ + \sin 200^\circ = 0 $ $ \sin(4x + 100^\circ) = -2\sin 150^\circ \cos 50^\circ $ $ \sin(4x + 100^\circ) = -\cos 50^\circ = \sin(-40^\circ) $ $ 4x + 100^\circ = n\pi + (-1)^n(-40^\circ) $ $ x = \frac{n\pi + (-1)^n(40^\circ) - 100^\circ}{4} $ $ \Rightarrow x = 30^\circ,; 55^\circ,; 120^\circ,; 145^\circ \text{ in } (0,\pi) $ $ \therefore $ no. of solutions $ = 4 $

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