Trigonometry Previous Year Questions (JEE Main 2026)

Use the identity for the difference of tangents. For any angle $\theta$,

$$\tan 3\theta - \tan\theta = \frac{\sin 3\theta\cos\theta - \cos 3\theta\sin\theta}{\cos 3\theta\cos\theta} = \frac{\sin 2\theta}{\cos 3\theta\cos\theta} = \frac{2\sin\theta\cos\theta}{\cos 3\theta\cos\theta} = \frac{2\sin\theta}{\cos 3\theta}.$$

Hence each term of $A$ has the form

$$\frac{\sin\theta}{\cos 3\theta} = \tfrac12\left(\tan 3\theta - \tan\theta\right).$$

Applying this with $\theta = 3^\circ, 9^\circ, 27^\circ$:

$$A = \tfrac12\Big[(\tan 9^\circ - \tan 3^\circ) + (\tan 27^\circ - \tan 9^\circ) + (\tan 81^\circ - \tan 27^\circ)\Big].$$

The sum telescopes:

$$A = \tfrac12\left(\tan 81^\circ - \tan 3^\circ\right) = \frac{B}{2}.$$

Therefore

$$\frac{B}{A} = 2.$$

Answer: 2

Let $u = 1-\alpha$ and $v = 1-\beta$. Taking $\tan$ of both sides,

$$\frac{u+v}{1-uv} = \tan\frac{\pi}{4} = 1 \quad\Longrightarrow\quad u+v = 1 - uv.$$

Compute each piece. First,

$$u+v = 2 - (\alpha+\beta).$$

Next, since $\alpha\beta = \alpha\cdot\dfrac{1}{3\alpha} = \dfrac13$,

$$uv = (1-\alpha)(1-\beta) = 1 - (\alpha+\beta) + \alpha\beta = \frac43 - (\alpha+\beta).$$

So $1 - uv = (\alpha+\beta) - \dfrac13$. Substituting into $u+v = 1-uv$:

$$2 - (\alpha+\beta) = (\alpha+\beta) - \frac13 \quad\Longrightarrow\quad 2(\alpha+\beta) = \frac{7}{3} \quad\Longrightarrow\quad \alpha+\beta = \frac{7}{6}.$$

(For $\alpha+\beta = \tfrac76$ with $\alpha\beta = \tfrac13$, we get $\alpha = \tfrac12,\ \beta = \tfrac23$, which satisfies $0<\alpha<1$.)

Therefore

$$6(\alpha+\beta) = 6\cdot\frac{7}{6} = 7.$$

Answer: B

Rewrite the equation as a single sinusoid:

$$\sqrt{3}\sin\theta - \cos\theta = 1 \quad\Longrightarrow\quad 2\sin\!\left(\theta - \frac{\pi}{6}\right) = 1 \quad\Longrightarrow\quad \sin\!\left(\theta - \frac{\pi}{6}\right) = \frac12.$$

Thus

$$\theta - \frac{\pi}{6} = \frac{\pi}{6} + 2k\pi \quad\text{or}\quad \theta - \frac{\pi}{6} = \frac{5\pi}{6} + 2k\pi,$$

giving

$$\theta = \frac{\pi}{3} + 2k\pi \quad\text{or}\quad \theta = \pi + 2k\pi.$$

Collecting the values in $(-2\pi, 2\pi)$:

$$\theta = \frac{\pi}{3},\ -\frac{5\pi}{3} \quad\text{and}\quad \theta = \pi,\ -\pi.$$

Summing,

$$\sum_{\theta\in S}\theta = \frac{\pi}{3} - \frac{5\pi}{3} + \pi - \pi = -\frac{4\pi}{3}.$$

Answer: B

Convert to degrees: $\dfrac{\pi}{18}=10^\circ,\ \dfrac{5\pi}{18}=50^\circ,\ \dfrac{7\pi}{18}=70^\circ$, so

$$K = \sin 10^\circ\,\sin 50^\circ\,\sin 70^\circ.$$

Using $\sin 50^\circ = \cos 40^\circ$, $\sin 70^\circ = \cos 20^\circ$, $\sin 10^\circ = \cos 80^\circ$,

$$K = \cos 20^\circ\,\cos 40^\circ\,\cos 80^\circ = \frac18,$$

the standard product (multiply and divide by $\sin 20^\circ$ and use $\sin 2x = 2\sin x\cos x$ repeatedly).

Now

$$\frac{10K\pi}{3} = \frac{10\pi}{3}\cdot\frac18 = \frac{10\pi}{24} = \frac{5\pi}{12} = 75^\circ.$$

Therefore

$$\sin 75^\circ = \sin(45^\circ + 30^\circ) = \frac{\sqrt3+1}{2\sqrt2}.$$

Answer: A

Simplify the expression:

$$\sin x(\sin x + \cos x) = \sin^2 x + \sin x\cos x = \frac{1-\cos 2x}{2} + \frac{\sin 2x}{2} = \frac12 + \frac{1}{\sqrt2}\sin\!\left(2x - \frac{\pi}{4}\right).$$

Its range is $\left[\tfrac12 - \tfrac{1}{\sqrt2},\ \tfrac12 + \tfrac{1}{\sqrt2}\right] \approx [-0.207,\ 1.207]$, so the only integer values of $a$ are $a = 0$ and $a = 1$.

Case $a = 0$: $\sin x(\sin x + \cos x) = 0$.

• $\sin x = 0 \Rightarrow x = -\pi,\ 0,\ \pi$ (3 values),
• $\sin x + \cos x = 0 \Rightarrow \tan x = -1 \Rightarrow x = -\tfrac{\pi}{4},\ \tfrac{3\pi}{4}$ (2 values).

That gives 5 distinct solutions.

Case $a = 1$: $\dfrac{1}{\sqrt2}\sin\!\left(2x - \tfrac{\pi}{4}\right) = \tfrac12 \Rightarrow \sin\!\left(2x - \tfrac{\pi}{4}\right) = \tfrac{1}{\sqrt2}$. With $x\in[-\pi,\pi]$, the argument $2x-\tfrac{\pi}{4}$ ranges over $\left[-\tfrac{9\pi}{4}, \tfrac{7\pi}{4}\right]$, and $\sin(\cdot)=\tfrac{1}{\sqrt2}$ yields 4 values: $x = -\tfrac{3\pi}{4}, -\tfrac{\pi}{2}, \tfrac{\pi}{4}, \tfrac{\pi}{2}$.

Total:

$$n(S) = 5 + 4 = 9.$$

Answer: D

Put $t = \dfrac{x}{2}$, so $t \in \left[0, \dfrac{\pi}{2}\right]$ and we maximize

$$f(t) = 16\sin t\cos^3 t.$$

Differentiate:

$$f'(t) = 16\left(\cos^4 t - 3\sin^2 t\cos^2 t\right) = 16\cos^2 t\left(\cos^2 t - 3\sin^2 t\right).$$

Setting $f'(t)=0$ on $\left(0,\tfrac{\pi}{2}\right)$ gives $\cos^2 t = 3\sin^2 t$, i.e. $\tan^2 t = \tfrac13$, so $t = \dfrac{\pi}{6}$.

At $t = \dfrac{\pi}{6}$: $\sin t = \dfrac12,\ \cos t = \dfrac{\sqrt3}{2}$, hence

$$f\!\left(\tfrac{\pi}{6}\right) = 16\cdot\frac12\cdot\left(\frac{\sqrt3}{2}\right)^3 = 16\cdot\frac12\cdot\frac{3\sqrt3}{8} = 3\sqrt3.$$

This exceeds the endpoint values $f(0)=0$ and $f\!\left(\tfrac{\pi}{2}\right)=0$, so the maximum is $3\sqrt3$.

Answer: B

Factor the difference of eighth powers:

$$\cos^8\theta - \sin^8\theta = \left(\cos^4\theta - \sin^4\theta\right)\left(\cos^4\theta + \sin^4\theta\right).$$

Now $\cos^4\theta - \sin^4\theta = \cos 2\theta$ and $\cos^4\theta + \sin^4\theta = 1 - 2\sin^2\theta\cos^2\theta = 1 - \tfrac12\sin^2 2\theta$. So

$$2(\cos^8\theta - \sin^8\theta)\sec 2\theta = 2\cos 2\theta\left(1 - \tfrac12\sin^2 2\theta\right)\cdot\frac{1}{\cos 2\theta} = 2 - \sin^2 2\theta.$$

The constraint $\tan^2\theta \neq 1$ excludes exactly the angles where $\cos 2\theta = 0$, i.e. where $\sin^2 2\theta = 1$. Thus on $P$,

$$\sin^2 2\theta \in [0, 1) \quad\Longrightarrow\quad 2 - \sin^2 2\theta \in (1, 2].$$

We need $a^2 \in (1, 2]$ for some integer $a$. But $a^2 = 1$ is excluded (open endpoint) and the next square $a^2 = 4$ lies above $2$; no perfect square lies in $(1, 2]$.

Hence no integer $a$ works:

$$n(S) = 0.$$

Answer: A

Left side. If $\phi = \tan^{-1}(x\sqrt2)$ then $\tan\phi = x\sqrt2$, so

$$\sin\phi = \frac{x\sqrt2}{\sqrt{1 + 2x^2}}.$$

Right side. Let $\psi = \sin^{-1}\sqrt{1-x^2}$, so $\sin\psi = \sqrt{1-x^2}$ and (since $x\in(0,1)$, $\psi\in(0,\tfrac{\pi}{2})$) $\cos\psi = \sqrt{1-(1-x^2)} = x$. Then

$$\cot\psi = \frac{x}{\sqrt{1-x^2}}.$$

Equate and cancel $x>0$:

$$\frac{\sqrt2}{\sqrt{1+2x^2}} = \frac{1}{\sqrt{1-x^2}} \quad\Longrightarrow\quad 2(1-x^2) = 1 + 2x^2.$$

Solving:

$$2 - 2x^2 = 1 + 2x^2 \quad\Longrightarrow\quad 4x^2 = 1 \quad\Longrightarrow\quad x = \frac12.$$

Answer: A

By Vieta’s formulas, $\tan A + \tan B = 2$ and $\tan A\tan B = -5$. Then

$$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B} = \frac{2}{1-(-5)} = \frac{2}{6} = \frac13.$$

Since $\tan A\tan B = -5 < 0$, one of $A, B$ is positive and the other negative, and numerically $A+B$ is a small positive angle in the first quadrant, so $\cos(A+B) > 0$. With $\tan(A+B)=\tfrac13$,

$$\cos(A+B) = \frac{3}{\sqrt{10}}.$$

Using the half-angle identity $20\sin^2\!\left(\tfrac{A+B}{2}\right) = 10\big(1 - \cos(A+B)\big)$:

$$20\sin^2\!\left(\frac{A+B}{2}\right) = 10\left(1 - \frac{3}{\sqrt{10}}\right) = 10 - \frac{30}{\sqrt{10}} = 10 - 3\sqrt{10}.$$

Answer: C

Write everything in terms of $\cos x$:

$$3\sin^2 x + 12\cos x - 3 = 3(1 - \cos^2 x) + 12\cos x - 3 = -3\cos^2 x + 12\cos x.$$

Let $c = \cos x \in [-1, 1]$ and $g(c) = -3c^2 + 12c = -3(c-2)^2 + 12$. On $[-1,1]$ the vertex $c=2$ lies to the right, so $g$ is increasing; hence

$$g(-1) = -15,\qquad g(1) = 9,\qquad g(c) \in [-15, 9].$$

The equation has a solution exactly when $p \in [-15, 9]$. Summing all integers there:

$$\sum_{p=-15}^{9} p = \frac{(-15 + 9)\cdot 25}{2} = \frac{-6\cdot 25}{2} = -75.$$

Answer: C

Note that $2^{2p-1} = 2^{p}\cdot 2^{p-1}$ and $2^{p} - 2^{p-1} = 2^{p-1}$, so each term telescopes:

$$\tan^{-1}\!\left(\frac{2^{p-1}}{1 + 2^{2p-1}}\right) = \tan^{-1}\!\left(\frac{2^{p} - 2^{p-1}}{1 + 2^{p}\cdot 2^{p-1}}\right) = \tan^{-1} 2^{p} - \tan^{-1} 2^{p-1}.$$

Summing from $p=1$ to $11$:

$$\sum_{p=1}^{11}\left(\tan^{-1} 2^{p} - \tan^{-1} 2^{p-1}\right) = \tan^{-1} 2^{11} - \tan^{-1} 2^{0} = \tan^{-1} 2048 - \frac{\pi}{4}.$$

Therefore

$$\alpha = \frac{\pi}{4} + \tan^{-1} 2048 - \frac{\pi}{4} = \tan^{-1} 2048,$$

and

$$\tan\alpha = 2048.$$

Answer: 2048

Convert products to sums using $2\cos A\cos B = \cos(A-B) + \cos(A+B)$:

$$2\cos\theta\cos\frac{5\theta}{2} = \cos\frac{3\theta}{2} + \cos\frac{7\theta}{2}, \qquad 2\cos 7\theta\cos\frac{7\theta}{2} = \cos\frac{13\theta}{2} + \cos\frac{15\theta}{2}.$$

So the equation becomes

$$\cos\frac{3\theta}{2} + \cos\frac{7\theta}{2} = \cos\frac{13\theta}{2} + \cos\frac{15\theta}{2}.$$

Substitute $\theta = 2\phi$ (so $\theta \in [-\pi, \pi] \iff \phi \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$). The arguments become integer multiples of $\phi$:

$$\cos 3\phi + \cos 7\phi = \cos 13\phi + \cos 15\phi.$$

Applying $2\cos A\cos B$ once more, this is $2\cos 2\phi\cos 5\phi = 2\cos 14\phi\cos 7\phi$, i.e. $\cos 3\phi + \cos 7\phi = \cos 7\phi + \cos 21\phi$. The $\cos 7\phi$ terms cancel, leaving the clean equation

$$\boxed{\cos 3\phi = \cos 21\phi}.$$

Solve using $\cos X = \cos Y \iff X = 2n\pi \pm Y$:

$$21\phi = 2n\pi + 3\phi \implies \phi = \frac{n\pi}{9}, \qquad\qquad 21\phi = 2n\pi - 3\phi \implies \phi = \frac{n\pi}{12}.$$

Count in $\phi \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$:

• $\phi = \dfrac{n\pi}{9}$ needs $-\tfrac12 \le \tfrac{n}{9} \le \tfrac12$, i.e. $n = -4, \dots, 4$ — that is 9 values.
• $\phi = \dfrac{n\pi}{12}$ needs $-\tfrac12 \le \tfrac{n}{12} \le \tfrac12$, i.e. $n = -6, \dots, 6$ — that is 13 values.

Remove overlaps. Common values satisfy $\tfrac{n}{9} = \tfrac{m}{12}$, giving $\phi = 0, \pm\tfrac{\pi}{3}$ (i.e. $\theta = 0, \pm\tfrac{2\pi}{3}$) — 3 repeats.

$$n(S) = 9 + 13 - 3 = 19.$$

Answer: 19

Statement II. Since $\dfrac{6}{11} > \dfrac{1}{\sqrt2}$? Check the size directly: $\sin^{-1}\dfrac{6}{11} \approx 33.06^\circ$, so

$$\alpha = 3\sin^{-1}\frac{6}{11} \approx 99.2^\circ.$$

Because $90^\circ < \alpha < 180^\circ$, we get $\cos\alpha < 0$. Statement II is true.

Statement I. Also $\cos^{-1}\dfrac{4}{9} \approx 63.6^\circ$, so

$$\beta = 3\cos^{-1}\frac{4}{9} \approx 190.8^\circ,\qquad \alpha + \beta \approx 290.0^\circ.$$

An angle near $290^\circ$ lies in the fourth quadrant, where cosine is positive, so $\cos(\alpha+\beta) > 0$. Statement I is true.

Both statements are true.

Answer: A