Trigonometry-JEE-Main-PYQ’s

Mathematics Questions

If $A = \sin^2{x} + \cos^4{x},$ then for all real $x$ :

(A) $\frac{13}{16} \leq A \leq 1$

(B) $1 \leq A \leq 2$

(D) $\frac{3}{4} \leq A \leq 1$

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Let $\cos(\alpha + \beta) = \frac{4}{5}$ and $\sin(\alpha - \beta) = \frac{5}{13}$ , where $0 \leq \alpha, \beta \leq \frac{\pi}{4}$ . Then $\tan 2\alpha =$

(A) $\frac{56}{33}$

(B) $\frac{19}{12}$

(D) $\frac{25}{16}$

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Let $A$ and $B$ denote the statements

$A: \cos \alpha + \cos \beta + \cos \gamma = 0$

$B: \sin \alpha + \sin \beta + \sin \gamma = 0$

If $\cos (\beta - \gamma) + \cos (\gamma - \alpha) + \cos (\alpha - \beta) = -\frac{3}{2},$ then:

(A) $A \text{ is false and } B \text{ is true}$

(B) both $A$ and $B$ are true

(D) $A \text{ is true and } B \text{ is false}$

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If $0 < x < \pi$ and $\cos x + \sin x = \frac{1}{2},$ then $\tan x$ is:

(A) $\frac{1 - \sqrt{7}}{4}$

(B) $\frac{4 - \sqrt{7}}{3}$

(D) $\frac{1 + \sqrt{7}}{4}$

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If $u = \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta} + \sqrt{a^2 \sin^2 \theta + b^2 \cos^2 \theta},$ then the difference between the maximum and minimum values of $u^2$ is given by:

(A) $(a - b)^2$

(B) $2\sqrt{a^2 + b^2}$

(D) $2(a^2 + b^2)$

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Let $\alpha, \beta$ be such that $\pi < \alpha - \beta < 3\pi$ .

If $\sin \alpha + \sin \beta = -\frac{21}{65} \quad \text{and} \quad \cos \alpha + \cos \beta = -\frac{27}{65},$ then the value of $\cos \frac{\alpha - \beta}{2}$ is:

(A) $-\frac{6}{65}$

(B) $\frac{3}{\sqrt{130}}$

(D) $-\frac{3}{\sqrt{130}}$

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Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$ , where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$ . If $\tan 2\alpha = \frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}$ , $r, s \in N$ , then $r + s$ is equal to.

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If $\frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}$ , where $\alpha, \beta \in \mathbb{N}$ , then $\alpha + \beta$ is equal to.

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Let the set of all $a \in \mathbb{R}$ such that the equation $\cos 2x + a \sin x = 2a - 7$ has a solution be $[p, q]$ , and $r = \tan 9^\circ - \tan 27^\circ - \frac{1}{\cot 63^\circ} + \tan 81^\circ,$ then $pqr$ is equal to

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The value of $\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ$ is

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Trigonometry-JEE-Main-PYQ’s

Mathematics Questions

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