Trigonometry – Trigonometric Identities – JEE Main 28 Jan 2026 Shift 1

Solution:

We are given the equation:
$$\frac{\tan(A-B)}{\tan A} + \frac{\sin^2 C}{\sin^2 A} = 1$$

First, we use the formula for $\tan(A-B)$ and express $\sin^2 \theta$ in terms of $\tan \theta$ using $\sin^2 \theta = \frac{\tan^2 \theta}{1+\tan^2 \theta}$.

Let $\tan A = x$, $\tan B = y$, and $\tan C = z$.
Then the term $\frac{\sin^2 C}{\sin^2 A}$ becomes:
$$\frac{\sin^2 C}{\sin^2 A} = \frac{\frac{\tan^2 C}{1+\tan^2 C}}{\frac{\tan^2 A}{1+\tan^2 A}} = \frac{\frac{z^2}{1+z^2}}{\frac{x^2}{1+x^2}} = \frac{z^2(1+x^2)}{x^2(1+z^2)}$$

Now substitute this back into the original equation:
$$\frac{1}{x}\left(\frac{x-y}{1+xy}\right) + \frac{z^2(1+x^2)}{x^2(1+z^2)} = 1$$

Multiply the entire equation by $x^2(1+xy)(1+z^2)$ to remove fractions:
$$x(x-y)(1+z^2) + z^2(1+x^2)(1+xy) = x^2(1+xy)(1+z^2)$$

Expand the terms:
$$(x^2-xy)(1+z^2) + (z^2+x^2z^2)(1+xy) = (x^2+x^3y)(1+z^2)$$
$$x^2+x^2z^2-xy-xyz^2 + z^2+xyz^2+x^2z^2+x^3yz^2 = x^2+x^2z^2+x^3y+x^3yz^2$$

Simplify by canceling common terms ($x^2, x^2z^2, x^3yz^2$) from both sides:
$$-xy – xyz^2 + z^2 + xyz^2 + x^2z^2 = x^3y$$
$$-xy + z^2 + x^2z^2 = x^3y$$
$$z^2(1+x^2) = x^3y + xy$$
$$z^2(1+x^2) = xy(x^2+1)$$

Since $A \in (0, \pi/2)$, $x = \tan A > 0$, so $x^2+1 \neq 0$. We can divide by $(1+x^2)$:
$$z^2 = xy$$

Substituting back the tangent functions:
$$\tan^2 C = \tan A \cdot \tan B$$

This satisfies the condition for a Geometric Progression where the middle term is $\tan C$.
Therefore, $\tan A, \tan C, \tan B$ are in G.P.

Ans. (1)