AMU MCA Trigonometry Previous Year Questions (PYQs)
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Showing 9 questions
The equation $\sin x + \sin y + \sin z = -3$ for $0 \le x \le 2\pi,; 0 \le y \le 2\pi,; 0 \le z \le 2\pi$ has
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Solution
Let $A, B$ and $C$ are the angles of a plain triangle and $\tan\left(\dfrac{A}{2}\right)=\dfrac{1}{3},; \tan\left(\dfrac{B}{2}\right)=\dfrac{2}{3}$. Then $\tan\left(\dfrac{C}{2}\right)$ is equal to
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Solution
If $\alpha, \beta; (\alpha \ne \beta)$ satisfies the equation $a\cos\theta + b\sin\theta = c$, then the value of $\tan\left(\dfrac{\alpha + \beta}{2}\right)$ is
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Solution
Let $0<\alpha<\pi,\ 0<\beta<\pi$ and $\cos\alpha+\cos\beta-\cos(\alpha+\beta)=\frac{3}{2}$. Then the relation between $\alpha$ and $\beta$ will be
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Solution
Let $\frac{\sin(\theta-\alpha)}{\sin(\theta-\beta)}=\frac{a}{b},\ \frac{\cos(\theta-\alpha)}{\cos(\theta-\beta)}=\frac{c}{d}$. Then the value of $\cos(\alpha-\beta)$ equals
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Solution
If $\sin(\alpha+\beta)=1,\ \sin(\alpha-\beta)=\frac{1}{2},\ \alpha,\beta\in\left[0,\frac{\pi}{2}\right]$ then the value of $\tan(\alpha+2\beta)\tan(2\alpha+\beta)$ is
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Solution
$ \sin(\alpha+\beta)=1 $
$ \Rightarrow \alpha+\beta=\frac{\pi}{2} $
$ \sin(\alpha-\beta)=\frac{1}{2} $
$ \Rightarrow \alpha-\beta=\frac{\pi}{6} $
Solving,
$ \alpha=\frac{\pi}{3},\ \beta=\frac{\pi}{6} $
$ \tan(\alpha+2\beta)\tan(2\alpha+\beta) $
$ =\tan\left(\frac{\pi}{3}+\frac{\pi}{3}\right)\tan\left(\frac{2\pi}{3}+\frac{\pi}{6}\right) $
$ =\tan\left(\frac{2\pi}{3}\right)\tan\left(\frac{5\pi}{6}\right) $
$ =(-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)=1 $
The value of $\frac{\sin^3 3\theta}{\sin^2 \theta} - \frac{\cos^2 3\theta}{\cos^2 \theta}$ is equal to
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Solution
Use identities and simplify ⇒
Expression reduces to $8\cos 2\theta$
Final Answer: $\boxed{8\cos 2\theta}$
The value of the expression
$1 - \frac{\sin^2 y}{1+\cos y} + \frac{1+\cos y}{\sin y} - \frac{\sin y}{1-\cos y}$ is equal to
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Solution
Use identities:
$\frac{\sin^2 y}{1+\cos y} = 1-\cos y$
$\frac{\sin y}{1-\cos y} = \frac{1+\cos y}{\sin y}$
Terms cancel ⇒ result = $\cos y$
Final Answer: $\boxed{\cos y}$
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