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
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Correct Answer: n2
Solution : Given that $\frac{\cos\alpha }{\cos\beta }=m$ and $\frac{\cos\alpha }{\sin\beta }=n$, we can express $\cos\: \alpha$ in terms of $m$ and $n$ as follows: $⇒\cos\alpha = m×cos\beta =n× \sin\beta$ Squaring both sides, we get: $⇒(m × \cos\beta)^2 = (n × \sin\beta)^2$ This simplifies to: $⇒m^2 × \cos^2\beta = n^2 × (1 - \cos^2\beta)$ Rearranging terms, we get: $⇒m^2 × \cos^2\beta + n^2 × \cos^2\beta = n^2$ Factoring out $cos^2\beta$, we get: $⇒(m^2 + n^2) × \cos^2\beta = n^2$ Hence, the correct answer is n2.
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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 $\alpha+\beta=90^{\circ}$ 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: $\frac{4}{3}$
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 $\frac{\cos \beta}{\sec \alpha}=15$ 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: $-\frac{481}{256}$
Question : Simplify the following. $\frac{\sin^3 \alpha+\cos^3 \alpha}{\sin \alpha+\cos \alpha}$
Option 1: $1+\sin \alpha \cos \alpha$
Option 2: $\tan \alpha$
Option 3: $1-\sin \alpha \cos \alpha$
Option 4: $\sec \alpha$
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