JEE MAIN Trigonometry Previous Year Questions (PYQs) – Page 6 of 8
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Let $f(\theta)=3\big(\sin^{4}\!\left(\tfrac{3\pi}{2}-\theta\right)+\sin^{4}\!(3\pi+\theta)\big)-2\big(1-\sin^{2}2\theta\big)$ and
$S=\left\{\theta\in[0,\pi]:\, f'(\theta)=-\dfrac{\sqrt{3}}{2}\right\}$.
If $4\beta=\displaystyle\sum_{\theta\in S}\theta$, then $f(\beta)$ is equal to:
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Solution
If $ \frac{\tan(A - B)}{\tan A} + \frac{\sin^2 C}{\sin^2 A} = 1 $, $ A, B, C \in \left( 0, \frac{\pi}{2} \right) $, then
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Solution
$ \frac{\tan A - \tan B}{(1 + \tan A \tan B)\tan A} + \frac{1 + \cot^2 A}{1 + \cot^2 C} = 1 $
Put $\tan A = x,; \tan B = y,; \tan C = z$
$ \Rightarrow \frac{x - y}{(1 + xy)x} + \frac{(x^2 + 1)z^2}{x^2(z^2 + 1)} = 1 $
$ \Rightarrow x(x - y)(z^2 + 1) + z^2(1 + x^2)(1 + xy) = (1 + xy)x^2(1 + z^2) $
after solving we get
$ z^2 = xy \quad \therefore 1 + x^2 \ne 0 $
$ \therefore \tan^2 C = \tan A \cdot \tan B $
$ \Rightarrow \tan A,; \tan C,; \tan B$ are in G.P.
If 15sin4$\alpha$ + 10cos4$\alpha$ = 6, for some $\alpha$$\in$R, then the value of 27sec6$\alpha$ + 8cosec6$\alpha$ is equal to :
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Solution
The value of $\sin 10^{\circ},\sin 30^{\circ},\sin 50^{\circ},\sin 70^{\circ}$ is:
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Solution
If 15sin4$\alpha$ + 10cos4$\alpha$ = 6, for some $\alpha$$\in$R, then the value of 27sec6$\alpha$ + 8cosec6$\alpha$ is equal to :
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Solution
If $\theta \in \left[-\dfrac{7\pi}{6}, \dfrac{4\pi}{3}\right]$, then the number of solutions of
$\sqrt{3}\csc^2\theta - 2(\sqrt{3} - 1)\csc\theta - 4 = 0$
is equal to:
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Solution
Let $A = \{\theta \in (-\frac{\pi}{2}, \pi) : \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta} \text{ is purely imaginary}\}$.
Then the sum of the elements in $A$ is:
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Solution
The set of all values of \(\lambda\) for which the equation
\[
\cos^{2}(2x)-2\sin^{4}x-2\cos^{2}x=\lambda
\]
has a real solution \(x\), is:
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Solution
If $\tan A=\dfrac{1}{\sqrt{x^{2}+x+1}},\quad \tan B=\dfrac{\sqrt{x}}{\sqrt{x^{2}+x+1}}$ and
$\tan C=\big(x^{-3}+x^{-2}+x^{-1}\big)^{1/2}$ with $0
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Solution
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