Question : If $\sin\theta=\frac{9}{41}$, $0^{\circ}<\theta<90^{\circ}$. Then, what is the value of $\cot \theta $?
Option 1: $\frac{39}{9}$
Option 2: $\frac{47}{8}$
Option 3: $\frac{35}{8}$
Option 4: $\frac{40}{9}$
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: $\frac{40}{9}$
Solution : Given: $\sin\theta=\frac{9}{41}$ in the interval $0^{\circ}<\theta<90^{\circ}$. We know that $\mathrm{cosec}^2\theta–\cot^2\theta=1$. $\sin\theta=\frac{9}{41}$ ⇒ $\mathrm{cosec}\ \theta=\frac{41}{9}$ ⇒ $\cot^2\theta=\mathrm{cosec}^2 \ \theta–1$ ⇒ $\cot^2\theta=(\frac{41}{9})^2–1$ ⇒ $\cot^2 \theta=\frac{1681}{81}–1$ ⇒ $\cot^2\theta=\frac{1681–81}{81}$ ⇒ $\cot^2\theta=\frac{1600}{81}$ ⇒ $\cot\theta=\frac{40}{9}$ Hence, the correct answer is $\frac{40}{9}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
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 : $\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)} \times \frac{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}{\tan \theta+\cot \theta}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : The expression $\frac{\cos ^4 \theta-\sin ^4 \theta+2 \sin ^2 \theta+3}{(\operatorname{cosec} \theta+\cot \theta+1)(\operatorname{cosec} \theta-\cot \theta+1)-2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : If $5\cos\theta+12\sin\theta=13,\ 0^0<\theta<90^0$, then the value of $\sin\theta$ is:
Question : If $7\sin^{2}\theta+3\cos^{2}\theta=4$, and $0^{\circ}< \theta< 90^{\circ}$, then the value of $\tan\theta$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile