Question : If $\alpha+\beta=90^{\circ}$ and $\alpha=2 \beta$, then the value of $3 \cos ^2 \alpha-2 \sin ^2 \beta$ is equal to:
Option 1: $\frac{3}{4}$
Option 2: $\frac{3}{2}$
Option 3: $\frac{1}{4}$
Option 4: $\frac{4}{3}$
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Correct Answer: $\frac{1}{4}$
Solution : Given that $\alpha + \beta = 90^\circ$ and $\alpha = 2\beta$, Substitute $\alpha$ in terms of $\beta$ into the first equation, $⇒2\beta + \beta = 90^\circ$ $⇒\beta = 30^\circ$ $\therefore \alpha = 60^\circ$ Substitute these values into the expression $3\cos^2\alpha - 2\sin^2\beta$: $3\cos^2(60^\circ) - 2\sin^2(30^\circ) = 3\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right)^2 = 3\left(\frac{1}{4}\right) - 2\left(\frac{1}{4}\right) = \frac{1}{4}$ Hence, the correct answer is $ \frac{1}{4}$.
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Question : What is $\sin \alpha - \sin\beta$?
Option 1: $2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$
Option 2: $2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$
Option 3: $2 \cos \frac{\alpha-\beta}{2} \sin \frac{\alpha+\beta}{2}$
Option 4: $2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}$
Question : If $\frac{\cos \alpha}{\sin \beta} = 10$ and $\frac{\cos \alpha}{\cos \beta} = 11$, the value of $\cos ^2 \beta$ is:
Option 1: $\frac{121}{132}$
Option 2: $\frac{100}{221}$
Option 3: $\frac{88}{108}$
Option 4: $\frac{221}{121}$
Question : If $\alpha +\beta =90^\circ$ and $\alpha :\beta =2:1,$ then the ratio of $\cos \alpha$ to $\cos \beta$ is:
Option 1: $1:\sqrt{3}$
Option 2: $1:3$
Option 3: $1:\sqrt{2}$
Option 4: $1:2$
Question : If $\frac{\cos\alpha}{\cos\beta}= m$ and $\frac{\cos\alpha}{\sin\beta}= n$, then the value of $(m^{2}+n^{2})\cos^{2}\beta$ is:
Option 1: n2
Option 2: m2
Option 3: mn
Option 4: 1
Question : If $\frac{\cos \beta}{\sec \alpha}=15$ and $\frac{\sin \beta}{\sec \alpha}=16$, the value of $\sin ^2 \beta$ is:
Option 1: $\frac{256}{481}$
Option 2: $-\frac{256}{481}$
Option 3: $\frac{481}{256}$
Option 4: $-\frac{481}{256}$
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