JEE Main 2026 — 24 January, Evening Shift — Question 25

$\frac{\tan \left(\mathrm{x}+100^{\circ}\right)}{\tan \mathrm{x}}=\tan \left(\mathrm{x}+50^{\circ}\right) \tan \left(\mathrm{x}-50^{\circ}\right)$

$\frac{\sin \left(\mathrm{x}+100^{\circ}\right) \cos \mathrm{x}}{\cos \left(\mathrm{x}+100^{\circ}\right) \sin \mathrm{x}}=\frac{\sin \left(\mathrm{x}+50^{\circ}\right) \sin \left(\mathrm{x}-50^{\circ}\right)}{\cos \left(\mathrm{x}+50^{\circ}\right) \cos \left(\mathrm{x}-50^{\circ}\right)}$

Apply C & D $\frac{\sin \left(2 \mathrm{x}+100^{\circ}\right)}{\sin 100^{\circ}}=\frac{\cos 100^{\circ}}{-\cos 2 \mathrm{x}}$ $2 \sin \left(2 \mathrm{x}+100^{\circ}\right) \cos 2 \mathrm{x}+\sin 200^{\circ}=0$

$\sin \left(4 x+100^{\circ}\right)+\sin 100^{\circ}+\sin 200^{\circ}=0$

$\sin \left(4 \mathrm{x}+100^{\circ}\right)=-2 \sin 150^{\circ} \cos 50^{\circ}$

$\sin \left(4 \mathrm{x}+100^{\circ}\right)=-\cos 50^{\circ}=\sin \left(-40^{\circ}\right)$

$\therefore 4 \mathrm{x}+100^{\circ}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \cdot\left(-40^{\circ}\right)$

$\mathrm{x}=\frac{\mathrm{n} \pi+(-1)^{\mathrm{n}+1}\left(40^{\circ}\right)-100^{\circ}}{4}$

$\therefore \mathrm{x}=30^{\circ}, 55^{\circ}, 120^{\circ}, 145^{\circ}$ in $(0, \pi)$

∴ no. of solutions $=4$