Question

If \(\sin \theta = \frac{1}{\sqrt{11}}\text{, then } \cot \theta\) equals

• A. \(\frac{\sqrt{11}}{\sqrt{10}}\)
• B. \(\frac{\sqrt{10}}{\sqrt{11}}\)
• C. \(\sqrt{10}\)
• D. \(\sqrt{11}\)

Hint:

Using trigonometric identities is often faster:
\[ \cot^2 \theta = \csc^2 \theta - 1 \]
Since \(\csc \theta = \frac{1}{\sin \theta} = \sqrt{11}\):
\[ \cot^2 \theta = (\sqrt{11})^2 - 1 = 11 - 1 = 10 \]
\[ \cot \theta = \sqrt{10} \]
This identity-based approach helps avoid drawing triangles.

Answer 1

The Correct Option is C

Step 1: Set Up Using a Right Triangle.
We are given $\sin\theta = \dfrac{1}{\sqrt{11}}$. In a right triangle, $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$. So let opposite $= 1$ and hypotenuse $= \sqrt{11}$.
Step 2: Find the Adjacent Side Using Pythagoras.
\[ \text{adjacent}^2 = \text{hypotenuse}^2 - \text{opposite}^2 = 11 - 1 = 10 \] \[ \text{adjacent} = \sqrt{10} \]
Step 3: Recall the Definition of $\cot\theta$.
\[ \cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{\text{adjacent}}{\text{opposite}} \]
Step 4: Compute $\cot\theta$.
\[ \cot\theta = \frac{\sqrt{10}}{1} = \sqrt{10} \]
Step 5: Verify Using the Identity $\cot^2\theta = \csc^2\theta - 1$.
$\csc\theta = \sqrt{11}$, so $\cot^2\theta = 11 - 1 = 10$, giving $\cot\theta = \sqrt{10}$. Confirmed!
Step 6: Match with Options.
$\sqrt{10}$ corresponds to option (3).
\[ \boxed{\sqrt{10}} \]