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
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
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.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
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 $x=a(\sin\theta+\cos\theta), y=b(\sin\theta-\cos\theta)$, then the value of $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ is:
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 $\sin\theta+\sin^{2}\theta=1$, then the value of $\cos^{12}\theta+3\cos^{10}\theta+3\cos^{8}\theta+\cos^{6}\theta-1$ is:
Question : If $8 \cot \theta=6$, then the value of $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile