JEE MAIN Trigonometry Previous Year Questions (PYQs) – Page 5 of 8
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For $\alpha,\beta\in(0,\dfrac{\pi}{2})$, let $3\sin(\alpha+\beta)=2\sin(\alpha-\beta)$ and a real number $k$ be such that $\tan\alpha=k\tan\beta$.
Then, the value of $k$ is equal to:
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Solution
If $\cot^{-1}(\alpha) = \cot^{-1}(2) + \cot^{-1}(8) + \cot^{-1}(18) + \cot^{-1}(32) + \ldots \text{ (upto 100 terms)},$ then $\alpha$ is :
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Solution
$\alpha = \sin 36^\circ $ is a root of which of the following equation?
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Solution
If the value of $\dfrac{5\cos36^{\circ}+5\sin18^{\circ}}{5\cos36^{\circ}-3\sin18^{\circ}}$ is $\dfrac{a\sqrt{5}-b}{c}$, where $a,b,c$ are natural numbers and $\gcd(a,c)=1$, then $a+b+c$ is equal to:
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Solution
The largest value of $ n $, for which $ 40^n $ divides $ 60! $, is:
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Solution
$ 40^n = 2^{3n} \times 5^n $
$ E_2(60!) = \left[ \frac{60}{2} \right] + \left[ \frac{60}{2^2} \right] + \left[ \frac{60}{2^3} \right] + \left[ \frac{60}{2^4} \right] + \left[ \frac{60}{2^5} \right] $
$ = 30 + 15 + 7 + 3 + 1 = 56 $
$ E_5(60!) = \left[ \frac{60}{5} \right] + \left[ \frac{60}{5^2} \right] $
$ = 12 + 2 = 14 $
$ \therefore n = \min \left( \frac{56}{3}, 14 \right) = 14 $
Let $\lvert\cos\theta,\cos(60^\circ-\theta),\cos(60^\circ+\theta)\rvert\le \dfrac{1}{8},;\theta\in[0,2\pi]$.
Then the sum of all $\theta\in[0,2\pi]$ where $\cos 3\theta$ attains its maximum value is:
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Solution
The sum of the solutions $x \in \mathbb{R}$ of the equation
$\dfrac{3\cos 2x + \cos^3 2x}{\cos^6 x - \sin^6 x} = x^3 - x^2 + 6$
is:
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Solution
The value of $\cos^2 10^\circ - \cos 10^\circ \cos 50^\circ + \cos^2 50^\circ$ is
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Solution
If $\theta \in[-2 \pi, 2 \pi]$, then the number of solutions of $2 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0$, is equal to:
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Solution
The number of elements in the set
$ { x \in [0,180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ)\tan x \tan(x - 50^\circ) } $
is ______
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Solution
$ \frac{\tan(x + 100^\circ)}{\tan x} = \tan(x + 50^\circ)\tan(x - 50^\circ) $
$ \frac{\sin(x + 100^\circ)\cos x}{\cos(x + 100^\circ)\sin x} = \frac{\sin(x + 50^\circ)\sin(x - 50^\circ)}{\cos(x + 50^\circ)\cos(x - 50^\circ)} $
Apply $ C $ & $ D $
$ \frac{\sin(2x + 100^\circ)}{\sin 100^\circ} = \frac{\cos 100^\circ}{-\cos 2x} $
$ 2\sin(2x + 100^\circ)\cos 2x + \sin 100^\circ + \sin 200^\circ = 0 $
$ \sin(4x + 100^\circ) + \sin 100^\circ + \sin 200^\circ = 0 $
$ \sin(4x + 100^\circ) = -2\sin 150^\circ \cos 50^\circ $
$ \sin(4x + 100^\circ) = -\cos 50^\circ = \sin(-40^\circ) $
$ 4x + 100^\circ = n\pi + (-1)^n(-40^\circ) $
$ x = \frac{n\pi + (-1)^n(40^\circ) - 100^\circ}{4} $
$ \Rightarrow x = 30^\circ,; 55^\circ,; 120^\circ,; 145^\circ \text{ in } (0,\pi) $
$ \therefore $ no. of solutions $ = 4 $
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