JEE Main 2026 Jan 24 Shift 2, Mathematics Q25: Trigonometric Equations

Rearrange: $\frac{\tan (x+100^{\circ })}{\tan x}=\tan (x+50^{\circ })\tan (x-50^{\circ })$.

Write each side in $\sin /\cos$ and apply the identity $2\sin A\cos B=\sin (A+B)+\sin (A-B)$. After applying componendo and dividendo on the resulting ratio:

$\frac{\sin (2x+100^{\circ })}{\sin 100^{\circ }}=\frac{\cos 100^{\circ }}{-\cos 2x}$.

Cross-multiplying: $2\sin (2x+100^{\circ })\cos 2x+\sin 200^{\circ }=0$.

$\sin (4x+100^{\circ })+\sin 100^{\circ }+\sin 200^{\circ }=0\Rightarrow \sin (4x+100^{\circ })=-2\sin 150^{\circ }\cos 50^{\circ }=-\cos 50^{\circ }=\sin (-40^{\circ })$.

So $4x+100^{\circ }=n\pi +(-1{)}^n(-40^{\circ })$, giving $x=\frac{n\pi +(-1{)}^{n+1}(40^{\circ })-100^{\circ }}{4}$.

Checking $x$ in $(0^{\circ },180^{\circ })$ for integer $n$, the valid solutions are $x=30^{\circ },55^{\circ },120^{\circ },145^{\circ }$.

Number of elements $=4$.