JEE MAIN Trigonometry Previous Year Questions (PYQs) – Page 4 of 8
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If $\sin^{4}\alpha+4\cos^{4}\beta+2=4\sqrt{2},\sin\alpha\cos\beta;\ \alpha,\beta\in[0,\pi],$ then $\cos(\alpha+\beta)-\cos(\alpha-\beta)$ is equal to:
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Solution
The sum of the solutions $x \in \mathbb{R}$ of the equation $\frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6$ is
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Solution
If $5\left(\tan^{2}x-\cos^{2}x\right)=2\cos2x+9$,
then the value of $\cos4x$ is:
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Solution
If the equation cos4 $\theta $ + sin4 $\theta $ +$\lambda $= 0 has real solutions for $\theta $, then$\lambda $ lies in the interval :
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Solution
$16\sin (20^\circ )\sin (40^\circ )\sin (80^\circ )$ is equal to :
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Solution
The value of
$\dfrac{\sqrt{3}\csc20^\circ-\sec20^\circ}{\cos20^\circ\cos40^\circ\cos60^\circ\cos80^\circ}$
is equal to:
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Solution
$E=\frac{\sqrt{3}}{\sin20^\circ}-\frac{1}{\cos20^\circ}$
$=\frac{\sqrt{3}\cos20^\circ-\sin20^\circ}{\sin20^\circ\cos20^\circ}$
$=\frac{2\sin(60^\circ-20^\circ)}{2\sin20^\circ\cos20^\circ}$
$=\frac{\sin40^\circ}{\sin40^\circ}=1$
Hence total
$=64$
If $\alpha,\ -\tfrac{\pi}{2} < \alpha < \tfrac{\pi}{2}$ is the solution of
$4\cos\theta + 5\sin\theta = 1$,
then the value of $\tan\alpha$ is:
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Solution
If $\sin x=-\frac{3}{5}$, where $\pi< x <\frac{3 \pi}{2}$, then $80\left(\tan ^2 x-\cos x\right)$ is equal to
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Solution
If $2\sin^3x+\sin2x\cos x+4\sin x-4=0$ has exactly $3$ solutions in the interval $\left[0,\dfrac{n\pi}{2}\right],,n\in\mathbb N$, then the roots of the equation $x^2+nx+(n-3)=0$ belong to:
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Solution
The value of $\cos \left( {{{2\pi } \over 7}} \right) + \cos \left( {{{4\pi } \over 7}} \right) + \cos \left( {{{6\pi } \over 7}} \right)$ is equal to :
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Solution
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