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JEE Main 202624 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Trigonometric Equations

The number of elements in the set \ x ∈[0,180^ ]: (x+100^ )= (x+50^...

The number of elements in the set {x∈[0,180∘]:tan⁡(x+100∘)=tan⁡(x+50∘)tan⁡xtan⁡(x−50∘)}\left\{x \in\left[0,180^{\circ}\right]: \tan\left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right)\tan x\tan\left(x-50^{\circ}\right)\right\} is ____\_\_\_\_.

Correct answer

4

Step-by-step solution

tan⁡(x+100°)tan⁡x=tan⁡(x+50°)tan⁡(x−50°)\frac{\tan(x+100°)}{\tan x} = \tan(x+50°)\tan(x-50°)
sin⁡(x+100°)cos⁡xcos⁡(x+100°)sin⁡x=sin⁡(x+50°)sin⁡(x−50°)cos⁡(x+50°)cos⁡(x−50°)\frac{\sin(x+100°)\cos x}{\cos(x+100°)\sin x} = \frac{\sin(x+50°)\sin(x-50°)}{\cos(x+50°)\cos(x-50°)}
Applying Componendo & Dividendo:
sin⁡(2x+100°)sin⁡100°=cos⁡100°−cos⁡2x\frac{\sin(2x+100°)}{\sin 100°} = \frac{\cos 100°}{-\cos 2x}
2sin⁡(2x+100°)cos⁡2x+sin⁡200°=02\sin(2x+100°)\cos 2x + \sin 200° = 0
sin⁡(4x+100°)+sin⁡100°+sin⁡200°=0\sin(4x+100°) + \sin 100° + \sin 200° = 0
sin⁡(4x+100°)=−2sin⁡150°cos⁡50°\sin(4x+100°) = -2\sin 150° \cos 50°
sin⁡(4x+100°)=−cos⁡50°=sin⁡(−40°)\sin(4x+100°) = -\cos 50° = \sin(-40°)
∴4x+100°=nπ+(−1)n(−40°)\therefore 4x + 100° = n\pi + (-1)^n(-40°)
x=nπ+(−1)n+1(40°)−100°4x = \frac{n\pi + (-1)^{n+1}(40°) - 100°}{4}
∴x=30°,55°,120°,145°\therefore x = 30°, 55°, 120°, 145° in (0,π)(0, \pi)
∴\therefore Number of solutions =4= 4
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