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: $\frac{40}{9}$
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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 $\frac{40}{9}$.
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