Q49 / 75 · JEE Main 2026 (24 January Shift 2) · Mathematics
The number of elements in the set $\left\{x \in\left[0,180^{\circ}\right]: \tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan x \tan \left(x-50^{\circ}\right)\right\}$ is $\_\_\_\_$.
Correct answer: 4
Solution
$\frac{\tan(x+100°)}{\tan x} = \tan(x+50°)\tan(x-50°)$
$\frac{\sin(x+100°)\cos x}{\cos(x+100°)\sin x} = \frac{\sin(x+50°)\sin(x-50°)}{\cos(x+50°)\cos(x-50°)}$
Applying Componendo & Dividendo:
$\frac{\sin(2x+100°)}{\sin 100°} = \frac{\cos 100°}{-\cos 2x}$
$2\sin(2x+100°)\cos 2x + \sin 200° = 0$
$\sin(4x+100°) + \sin 100° + \sin 200° = 0$
$\sin(4x+100°) = -2\sin 150° \cos 50°$
$\sin(4x+100°) = -\cos 50° = \sin(-40°)$
$\therefore 4x + 100° = n\pi + (-1)^n(-40°)$
$x = \frac{n\pi + (-1)^{n+1}(40°) - 100°}{4}$
$\therefore x = 30°, 55°, 120°, 145°$ in $(0, \pi)$
$\therefore$ Number of solutions $= 4$
$\frac{\sin(x+100°)\cos x}{\cos(x+100°)\sin x} = \frac{\sin(x+50°)\sin(x-50°)}{\cos(x+50°)\cos(x-50°)}$
Applying Componendo & Dividendo:
$\frac{\sin(2x+100°)}{\sin 100°} = \frac{\cos 100°}{-\cos 2x}$
$2\sin(2x+100°)\cos 2x + \sin 200° = 0$
$\sin(4x+100°) + \sin 100° + \sin 200° = 0$
$\sin(4x+100°) = -2\sin 150° \cos 50°$
$\sin(4x+100°) = -\cos 50° = \sin(-40°)$
$\therefore 4x + 100° = n\pi + (-1)^n(-40°)$
$x = \frac{n\pi + (-1)^{n+1}(40°) - 100°}{4}$
$\therefore x = 30°, 55°, 120°, 145°$ in $(0, \pi)$
$\therefore$ Number of solutions $= 4$
Hence, the correct answer is 4.