NIMCET 2026 — Mathematics PYQ

To find which statement is incorrect, we evaluate the trigonometric relationships in a triangle.

1. Verification of Option (d)

This is the standard Law of Sines, which relates the side lengths to the sines of their opposite angles:

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$

This formula is fundamentally correct for all triangles.

2. Verification of Option (c)

Using the Law of Sines where $a=k\sin A$ and $b=k\sin B$ (where $k$ is the diameter of the circumcircle), we substitute into the expression:

$$\frac{a-b}{a+b}=\frac{k\sin A-k\sin B}{k\sin A+k\sin B}=\frac{\sin A-\sin B}{\sin A+\sin B}$$

This identity is correct.

3. Verification of Option (a)

Applying the sum-to-product trigonometric formulas to the result of option (c):

$$\frac{\sin A-\sin B}{\sin A+\sin B}=\frac{2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)}{2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}$$

By simplifying, we get:

$$\frac{a-b}{a+b}=\cot \left(\frac{A+B}{2}\right)\tan \left(\frac{A-B}{2}\right)$$

This is the standard Napier's Analogy (Law of Tangents), which is correct.

4. Verification of Option (b)

We know that in $△ABC$, $A+B+C=\pi$, therefore $\frac{A+B}{2}=\frac{\pi }{2}-\frac{C}{2}$.

Using the co-function identity:

$$\cot \left(\frac{A+B}{2}\right)=\cot \left(\frac{\pi }{2}-\frac{C}{2}\right)=\tan \left(\frac{C}{2}\right)$$

Substituting this into the correct formula from option (a):

$$\frac{a-b}{a+b}=\tan \left(\frac{C}{2}\right)\tan \left(\frac{A-B}{2}\right)$$

Comparing this to option (b), which gives $\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{C}{2}\right)}$, we see that the denominator in the option is $\tan \left(\frac{C}{2}\right)$ instead of $1/\tan \left(\frac{C}{2}\right)$ or $\cot \left(\frac{C}{2}\right)$.

Therefore, Option (b) is the incorrect statement.