Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
If $\sin\left(\frac{\pi}{18}\right)\sin\left(\frac{5\pi}{18}\right)\sin\left(\frac{7\pi}{18}\right) = K$, then the value of $\sin\left(\frac{10K\pi}{3}\right)$ is:
(1) $\frac{\sqrt{3}+1}{2\sqrt{2}}$
(2) $\frac{\sqrt{3}-1}{\sqrt{2}}$
(3) $\frac{\sqrt{3}}{2}$
(4) $\frac{1}{2}$
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Answer: Option (1)
Key idea: Product of sines at specific angles and compound angle formulas
Step 1: Recognize that $\sin\left(\frac{\pi}{18}\right)\sin\left(\frac{5\pi}{18}\right)\sin\left(\frac{7\pi}{18}\right) = \sin(10°)\sin(50°)\sin(70°)$
Step 2: Use the identity $\sin(x)\sin(60°-x)\sin(60°+x) = \frac{1}{4}\sin(3x)$ with $x=10°$ to get $K = \frac{1}{4}\sin(30°) = \frac{1}{8}$
If A = $\frac{\sin 3^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 27^\circ} + \frac{\sin 27^\circ}{\cos 81^\circ}$ and B = $\tan 81^\circ - \tan 3^\circ$, then $\frac{B}{A}$ is equal to ______.
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Answer: 2
Key idea: Telescoping series using trigonometric identities
Use the identity: $\frac{\sin\theta}{\cos 3\theta} = \frac{\tan 3\theta - \tan\theta}{2}$
Apply to first term: $\frac{\sin 3°}{\cos 9°} = \frac{\tan 9° - \tan 3°}{2}$
Apply to second term: $\frac{\sin 9°}{\cos 27°} = \frac{\tan 27° - \tan 9°}{2}$
Apply to third term: $\frac{\sin 27°}{\cos 81°} = \frac{\tan 81° - \tan 27°}{2}$
Sum the terms (telescoping): $A = \frac{\tan 81° - \tan 3°}{2} = \frac{B}{2}$
$\max_{0 \le x \le \pi} \left\{16\sin\left(\frac{x}{2}\right)\left|\cos^3\left(\frac{x}{2}\right)\right|\right\}$ is equal to:
(1) $\frac{3\sqrt{3}}{2}$
(2) $3\sqrt{3}$
(3) $4\sqrt{3}$
(4) $6\sqrt{3}$
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Answer: Option (2)
Key idea: Optimization of trigonometric functions using calculus
Substitute $t = \frac{x}{2}$ where $0 \le t \le \frac{\pi}{2}$. Since both $\sin(t)$ and $\cos(t)$ are non-negative in this range, the function becomes $f(t) = 16\sin(t)\cos^3(t)$
Set $f'(t) = 0$: Either $\cos(t) = 0$ or $\cos^2(t) = 3\sin^2(t)$
From $\cos^2(t) = 3\sin^2(t)$ and $\sin^2(t) + \cos^2(t) = 1$, we get $\cos^2(t) = \frac{3}{4}$, so $\cos(t) = \frac{\sqrt{3}}{2}$ and $\sin(t) = \frac{1}{2}$
Let $\tan A$, $\tan B$, where $A, B \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, be the roots of the quadratic equation $x^2 - 2x - 5 = 0$. Then $20\sin^2\left(\frac{A+B}{2}\right)$ is equal to:
(1) $10 + \sqrt{10}$
(2) $10 - 2\sqrt{10}$
(3) $10 - 3\sqrt{10}$
(4) $10 - \sqrt{10}$
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Answer: Option (3)
Key idea: Tangent addition formula, half-angle formula, and Vieta's formulas
From Vieta's formulas: $\tan A + \tan B = 2$ and $\tan A \cdot \tan B = -5$
Find $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{2}{1-(-5)} = \frac{1}{3}$
From $\tan(A+B) = \frac{1}{3}$, find $\sin(A+B) = \frac{1}{\sqrt{10}}$ and $\cos(A+B) = \frac{3}{\sqrt{10}}$
For the functions $f(\theta) = \alpha \tan^2\theta + \beta \cot^2\theta$, and $g(\theta) = \alpha \sin^2\theta + \beta \cos^2\theta$, $\alpha > \beta > 0$, let $\min_{0 < \theta < \frac{\pi}{2}} f(\theta) = \max_{0 < \theta < \pi} g(\theta)$. If the first term of a G.P. is $\left(\frac{\alpha}{2\beta}\right)$, its common ratio is $\left(\frac{2\beta}{\alpha}\right)$ and the sum of its first 10 terms is $\frac{m}{n} \cdot \gcd(m, n) = 1$, then $m + n$ is equal to _________.
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Answer: 1279
Key idea: Optimization of trigonometric functions using AM-GM inequality, geometric progression sum formula
Find $\min f(\theta)$ using AM-GM: $\alpha \tan^2\theta + \beta \cot^2\theta \geq 2\sqrt{\alpha\beta}$, so $\min f(\theta) = 2\sqrt{\alpha\beta}$
Find $\max g(\theta)$: Write $g(\theta) = \beta + (\alpha - \beta)\sin^2\theta$. Since $\alpha > \beta$, maximum occurs at $\sin^2\theta = 1$, giving $\max g(\theta) = \alpha$
The sum of all possible values of $\theta \in [0, 2\pi]$, for which the system of equations:
$x\cos 3\theta - 8y - 12z = 0$
$x\cos 2\theta + 3y + 3z = 0$
$x + y + 3z = 0$
has a non-trivial solution, is equal to:
(1) $\pi$
(2) $2\pi$
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Determinant condition for non-trivial solutions of homogeneous linear systems and solving trigonometric equations
FAQs
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How many questions come from Trigonometry (Ratios, Identities, Equations) in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 9 questions on it, about 1.0 per shift.
Where can I practise Trigonometry (Ratios, Identities, Equations) PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
