NIMCET Trigonometric Ratios & Identities: High-Yield Solved PYQs

Practicing Previous Year Questions (PYQs) is the single most effective way to crack NIMCET Mathematics. Over 40% of trigonometry questions in NIMCET are either direct conceptual repetitions or variants of past questions from NIMCET, JEE Main, and AIEEE papers.

In this practice guide, we have compiled 10 high-yield solved PYQs covering trigonometric ratios, compound angles, maximum/minimum values, and transformation identities with step-by-step derivations and speed shortcuts.

Solved PYQ 1: Trigonometric Identity Simplification

Question (NIMCET Past Paper): If $\sin\theta + \csc\theta = 2$, then the value of $\sin^{10}\theta + \csc^{10}\theta$ is:

• (A) $2^{10}$
• (B) $2$
• (C) $1$
• (D) $0$

Solution: Given $\sin\theta + \frac{1}{\sin\theta} = 2$. Let $x = \sin\theta$. Then $x + \frac{1}{x} = 2 \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1$. Thus, $\sin\theta = 1$ and $\csc\theta = 1$. Now, $\sin^{10}\theta + \csc^{10}\theta = (1)^{10} + (1)^{10} = 1 + 1 = 2$.

Correct Option: (B)

Solved PYQ 2: Difference of Fourth Powers

Question (NIMCET Past Paper): The value of $\cos^4\frac{\pi}{8} + \cos^4\frac{3\pi}{8} + \cos^4\frac{5\pi}{8} + \cos^4\frac{7\pi}{8}$ is:

• (A) $\frac{1}{2}$
• (B) $\frac{3}{2}$
• (C) $\frac{3}{4}$
• (D) $1$

Solution: Notice that $\frac{7\pi}{8} = \pi - \frac{\pi}{8} \implies \cos\frac{7\pi}{8} = -\cos\frac{\pi}{8} \implies \cos^4\frac{7\pi}{8} = \cos^4\frac{\pi}{8}$. Also, $\frac{5\pi}{8} = \pi - \frac{3\pi}{8} \implies \cos^4\frac{5\pi}{8} = \cos^4\frac{3\pi}{8}$.

So the expression reduces to: $2\left(\cos^4\frac{\pi}{8} + \cos^4\frac{3\pi}{8}\right)$ Since $\frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$, we have $\cos\frac{3\pi}{8} = \sin\frac{\pi}{8}$. $\text{Expression} = 2\left(\cos^4\frac{\pi}{8} + \sin^4\frac{\pi}{8}\right)$ Using $a^4 + b^4 = (a^2+b^2)^2 - 2a^2b^2$: $\cos^4\theta + \sin^4\theta = 1 - 2\sin^2\theta\cos^2\theta = 1 - \frac{1}{2}(2\sin\theta\cos\theta)^2 = 1 - \frac{1}{2}\sin^2(2\theta)$ Here $\theta = \frac{\pi}{8}$, so $2\theta = \frac{\pi}{4}$, and $\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$. $\text{Value} = 2\left(1 - \frac{1}{2}\left(\frac{1}{\sqrt{2}}\right)^2\right) = 2\left(1 - \frac{1}{4}\right) = 2\left(\frac{3}{4}\right) = \frac{3}{2}$

Correct Option: (B)

Solved PYQ 3: Reciprocal Trigonometric System

Question (NIMCET Past Paper): If $\sec\theta + \tan\theta = p$, then $\sin\theta$ is equal to:

• (A) $\frac{p^2+1}{p^2-1}$
• (B) $\frac{p^2-1}{p^2+1}$
• (C) $\frac{p}{p^2+1}$
• (D) $\frac{p^2-1}{2p}$

Solution: Using $\sec^2\theta - \tan^2\theta = 1$: $\sec\theta - \tan\theta = \frac{1}{p}$ Adding: $2\sec\theta = p + \frac{1}{p} = \frac{p^2+1}{p} \implies \sec\theta = \frac{p^2+1}{2p}$ Subtracting: $2\tan\theta = p - \frac{1}{p} = \frac{p^2-1}{p} \implies \tan\theta = \frac{p^2-1}{2p}$ Divide $\tan\theta$ by $\sec\theta$: $\sin\theta = \frac{\tan\theta}{\sec\theta} = \frac{\frac{p^2-1}{2p}}{\frac{p^2+1}{2p}} = \frac{p^2-1}{p^2+1}$

Correct Option: (B)

Solved PYQ 4: Product Series Evaluation

Question (NIMCET Past Paper): The value of $\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$ is:

• (A) $\frac{1}{8}$
• (B) $\frac{1}{16}$
• (C) $\frac{1}{32}$
• (D) $\frac{1}{4}$

Solution: We know $\cos 60^\circ = \frac{1}{2}$. The remaining product is $\cos 20^\circ \cos 40^\circ \cos 80^\circ$. Using the shortcut $\cos\theta \cos(60^\circ-\theta) \cos(60^\circ+\theta) = \frac{1}{4}\cos 3\theta$ with $\theta = 20^\circ$: $\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{4}\cos 60^\circ = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$ Total Value $= \frac{1}{2} \times \frac{1}{8} = \frac{1}{16}$.

Correct Option: (B)

Solved PYQ 5: Range & Maximum Value

Question (NIMCET Past Paper): The maximum value of $5\cos\theta + 3\cos\left(\theta + \frac{\pi}{3}\right) + 3$ is:

• (A) $5$
• (B) $7$
• (C) $10$
• (D) $8$

Solution: Expand $\cos\left(\theta + \frac{\pi}{3}\right)$: $\cos\left(\theta + \frac{\pi}{3}\right) = \cos\theta \cos\frac{\pi}{3} - \sin\theta \sin\frac{\pi}{3} = \frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta$ Substitute back: $5\cos\theta + 3\left(\frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta\right) + 3 = \left(5 + \frac{3}{2}\right)\cos\theta - \frac{3\sqrt{3}}{2}\sin\theta + 3$ $= \frac{13}{2}\cos\theta - \frac{3\sqrt{3}}{2}\sin\theta + 3$ This is of the form $a\cos\theta + b\sin\theta + c$. Max value $= \sqrt{a^2+b^2} + c$. $a^2 + b^2 = \left(\frac{13}{2}\right)^2 + \left(-\frac{3\sqrt{3}}{2}\right)^2 = \frac{169}{4} + \frac{27}{4} = \frac{196}{4} = 49$ $\sqrt{a^2+b^2} = \sqrt{49} = 7$ $\text{Maximum Value} = 7 + 3 = 10$

Correct Option: (C)

Solved PYQ 6: Tangent Addition in Triangles

Question (NIMCET Past Paper): In a triangle $ABC$, if $\tan A + \tan B + \tan C = 6$ and $\tan A \cdot \tan B = 2$, then $\tan C$ is equal to:

• (A) $1$
• (B) $2$
• (C) $3$
• (D) $4$

Solution: In any triangle $ABC$, $A + B + C = \pi$, so $\tan A + \tan B + \tan C = \tan A \tan B \tan C$. Given $\tan A + \tan B + \tan C = 6$, we have: $\tan A \tan B \tan C = 6$ Given $\tan A \tan B = 2$, substitute: $2 \cdot \tan C = 6 \implies \tan C = 3$

Correct Option: (C)

Solved PYQ 7: Quadratic Equation in Sine

Question (NIMCET Past Paper): If $2\sin^2\theta + 5\sin\theta - 3 = 0$, then the general solution for $\theta$ in $[0, 2\pi]$ is:

• (A) $\frac{\pi}{6}, \frac{5\pi}{6}$
• (B) $\frac{\pi}{3}, \frac{2\pi}{3}$
• (C) $\frac{\pi}{6}$
• (D) $\frac{7\pi}{6}, \frac{11\pi}{6}$

Solution: Let $x = \sin\theta$. $2x^2 + 5x - 3 = 0 \implies (2x - 1)(x + 3) = 0$ So $x = \frac{1}{2}$ or $x = -3$. Since $-1 \le \sin\theta \le 1$, $\sin\theta = -3$ is impossible. Thus $\sin\theta = \frac{1}{2}$. In $[0, 2\pi]$, $\theta = \frac{\pi}{6}$ and $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

Correct Option: (A)

Frequently Asked Questions (FAQ)

Q1: Are JEE Main questions repeated in NIMCET?

Yes! NIMCET frequently adapts AIEEE, JEE Main, and state MCA entrance questions. Practicing 10-year JEE Main trigonometry questions gives you a distinct edge.

Q2: How many questions in NIMCET Math come from Trigonometric Identities?

Direct questions on identities range between 3 to 5 questions (12 to 20 marks out of 600 total math marks).

Q3: What is the best strategy to tackle lengthy trigonometry options?

Use the Angle Substitution Method. Substitute $\theta = 0^\circ, 30^\circ, 45^\circ$, or $90^\circ$ into the question and options to eliminate choices instantly.

Q4: Which chapters in NIMCET require Trigonometry?

Calculus (limits, differentiation, integration), Vectors, Complex Numbers, and Coordinate Geometry all rely heavily on trigonometric transformations.