JEE Main 2026 — 28 January, Morning Shift — Question 4

$\frac{\tan A-\tan B}{(1+\tan A \tan B) \tan A}+\frac{1+\cot ^{2} A}{1+\cot ^{2} C}=1$

Put $\tan \mathrm{A}=\mathrm{x}, \tan \mathrm{B}=\mathrm{y}, \tan \mathrm{C}=\mathrm{z}$

$\therefore \frac{\mathrm{x}-\mathrm{y}}{(1+\mathrm{xy}) \mathrm{x}}+\frac{\left(\mathrm{x}^{2}+1\right) \mathrm{z}^{2}}{\mathrm{x}^{2}\left(\mathrm{z}^{2}+1\right)}=1$

$\therefore \mathrm{x}(\mathrm{x}-\mathrm{y})\left(\mathrm{z}^{2}+1\right)+\mathrm{z}^{2}\left(1+\mathrm{x}^{2}\right)(1+\mathrm{xy})$ $=(1+x y) x^{2}\left(1+z^{2}\right)$

after solving we get $z^{2}=x y$

$$\because 1+\mathrm{x}^{2} \neq 0$$

$\therefore \tan ^{2} \mathrm{C}=\tan \mathrm{A} \cdot \tan \mathrm{B}$

$\tan \mathrm{A}, \tan \mathrm{C}, \tan \mathrm{B}$ are in G.P.