Question : If $x=8(\sin \theta+\cos \theta)$ and $y=9(\sin \theta-\cos \theta)$, then the value of $\frac{x^2}{8^2}+\frac{y^2}{9^2}$ is:
Option 1: 4
Option 2: 6
Option 3: 8
Option 4: 2
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Correct Answer: 2
Solution : Given: $x=8(\sin \theta+\cos \theta)$ and $y=9(\sin \theta-\cos \theta)$ Now, $\frac{x^2}{8^2}+\frac{y^2}{9^2}$ = $\frac{[8(\sin \theta+\cos \theta)]^2}{8^2}+\frac{[9(\sin \theta-\cos \theta)]^2}{9^2}$ = $\frac{64(\sin^2 \theta+\cos^2 \theta+2\sin \theta\cos \theta)}{64}+\frac{81(\sin^2 \theta+\cos^2 \theta-2\sin \theta\cos \theta)}{81}$ = $1+2\sin \theta\cos \theta +1-2\sin \theta\cos \theta$ = $2$ Hence, the correct answer is 2.
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Question : If $x\sin^{3}\theta +y\cos^{3}\theta=\sin\theta\cos\theta$ and $x\sin\theta-y\cos\theta=0$, then the value of $\left ( x^{2}+y^{2} \right )$ equals:
Option 1: $1$
Option 2: $\frac{1}{2}$
Option 3: $\frac{3}2$
Option 4: $2$
Question : If $x=a(\sin\theta+\cos\theta), y=b(\sin\theta-\cos\theta)$, then the value of $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: –2
Question : If $\tan\theta=1$, then the value of $\frac{8\sin\theta\:+\:5\cos\theta}{\sin^{3}\theta\:–\:2\cos^{3}\theta\:+\:7\cos\theta}$ is:
Option 1: $2$
Option 2: $2\frac{1}{2}$
Option 3: $3$
Option 4: $\frac{4}{5}$
Question : If $\sin\theta+\sin^{2}\theta=1$, then the value of $\cos^{12}\theta+3\cos^{10}\theta+3\cos^{8}\theta+\cos^{6}\theta-1$ is:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 0
Question : If $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}=\frac{3}{2}$, then the value of $\sin ^4 \theta-\cos ^4 \theta$ is:
Option 1: $\frac{5}{12}$
Option 2: $\frac{12}{13}$
Option 3: $\frac{11}{12}$
Option 4: $\frac{5}{13}$
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