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
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Correct Answer: 2
Solution : Given: $x=a(\sin\theta+\cos\theta), y=b(\sin\theta-\cos\theta)$ So, $\frac{x}{a}=(\sin\theta+\cos\theta), \frac{y}{a}=(\sin\theta-\cos\theta)$ Now, $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ = $(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2$ = $\sin^2\theta+2×\sin\theta×\cos\theta+\cos^2\theta+\sin^2\theta-2×\sin\theta×\cos\theta+\cos^2\theta$ = $\sin^2\theta+\cos^2\theta+\sin^2\theta+\cos^2\theta$ = $1+1$ = $2$ 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:
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:
Question : If $\tan \theta=\frac{4}{3}$, then the value of $\frac{3\sin \theta+ 2\cos \theta}{3\sin \theta – 2 \cos \theta}$ is:
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:
Question : If $\sin (x+y) = \cos (x–y)$, then the value of $\cos^2 x$ is:
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