Question : If $7\sin^{2}\theta+3\cos^{2}\theta=4$, and $0^{\circ}< \theta< 90^{\circ}$, then the value of $\tan\theta$ is:
Option 1: $\frac{1}{\sqrt{2}}$
Option 2: $\frac{1}{\sqrt{3}}$
Option 3: $\sqrt{\frac{3}{2}}$
Option 4: $1$
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: $\frac{1}{\sqrt{3}}$
Solution : We know that $\sin^{2}\theta + \cos^{2}\theta = 1$ $7\sin^{2}\theta+3\cos^{2}\theta=4$ ⇒ $7\sin^{2}\theta+3(1-\sin^{2}\theta)=4$ ⇒ $4\sin^{2}\theta=1$ ⇒ $\sin\theta = \sqrt{\frac{1}{4}}$ or $\sin\theta = -\sqrt{\frac{1}{4}}$ Since $0^{\circ}< \theta< 90^{\circ}$, $\sin\theta$ is positive. ⇒ $\sin\theta = \frac{1}{2}$ ⇒ $\theta = 30^{\circ}$ ⇒ $\tan\theta = \tan30^{\circ} = \frac{1}{\sqrt{3}}$ Hence, the correct answer is $\frac{1}{\sqrt{3}}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : $\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)} \times \frac{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}{\tan \theta+\cot \theta}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : If $3 \tan \theta=2 \sqrt{3} \sin \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\operatorname{cosec}^2 2 \theta+\cot ^2 2 \theta}{\sin ^2 \theta+\tan ^2 2 \theta}$ is:
Question : If $\cos \left(2 \theta+54^{\circ}\right)=\sin \theta, 0^{\circ}<\left(2 \theta+54^{\circ}\right)<90^{\circ}$, then what is the value of $\frac{1}{\tan 5 \theta+\operatorname{cosec} \frac{5 \theta}{2}}$?
Question : If $1+2 \tan ^2 \theta+2 \sin \theta \sec ^2 \theta=\frac{a}{b}, 0^{\circ}<\theta<90^{\circ}$, then $\frac{a+b}{a-b}=?$
Question : The expression $\frac{(1-\sin \theta+\cos \theta)^2(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\sec \theta-\tan \theta)(\tan \theta+\cot \theta)}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile