NIMCET 2026 — Mathematics PYQ

To determine the relationship between $x$ and $y$, we simplify the given equation:

$$x-y\tan 35^{\circ }=y\tan 25^{\circ }+x\tan 25^{\circ }\tan 35^{\circ }$$

1. Grouping terms

Rearrange the equation to group $x$ and $y$ terms:

$$x-x\tan 25^{\circ }\tan 35^{\circ }=y\tan 25^{\circ }+y\tan 35^{\circ }$$

$$x(1-\tan 25^{\circ }\tan 35^{\circ })=y(\tan 25^{\circ }+\tan 35^{\circ })$$

2. Rewriting using Trigonometric Identities

Recall the addition formula for tangent: $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$.

Rearranging this, we get $\tan A+\tan B=\tan (A+B)(1-\tan A\tan B)$.

Substitute $A=25^{\circ }$ and $B=35^{\circ }$:

$$\tan 25^{\circ }+\tan 35^{\circ }=\tan (25^{\circ }+35^{\circ })(1-\tan 25^{\circ }\tan 35^{\circ })$$

$$\tan 25^{\circ }+\tan 35^{\circ }=\tan 60^{\circ }(1-\tan 25^{\circ }\tan 35^{\circ })$$

3. Substituting back into the main equation

$$x(1-\tan 25^{\circ }\tan 35^{\circ })=y[\tan 60^{\circ }(1-\tan 25^{\circ }\tan 35^{\circ })]$$

Assuming $(1-\tan 25^{\circ }\tan 35^{\circ })\ne 0$, we can divide both sides:

$$x=y\tan 60^{\circ }$$

Since $\tan 60^{\circ }=\sqrt{3}$:

$$x=y\sqrt{3}$$

Because \sqrt{3} \approx 1.732 > 1, for any positive $y$, $x$ will be greater than $y$ (x > y).

Conclusion: The correct relationship is (b) x > y.