Question

If value of \(\cot \theta\) is \(\sqrt{5}\), then \(\sin \theta\) equals

• A. \(\frac{1}{\sqrt{6}}\)
• B. \(\sqrt{6}\)
• C. \(\frac{\sqrt{5}}{6}\)
• D. \(\frac{1}{2}\)

Hint:

You can also use the trigonometric identity: \[ \csc^2 \theta = 1 + \cot^2 \theta \] \[ \csc^2 \theta = 1 + (\sqrt{5})^2 = 6 \implies \csc \theta = \sqrt{6} \] Since \(\sin \theta = \frac{1}{\csc \theta}\), we get \(\sin \theta = \frac{1}{\sqrt{6}}\). This is often faster!

Answer 1

The Correct Option is A

Step 1: Use the given value.
We are given \(\cot\theta = \sqrt{5}\), which means \(\frac{\cos\theta}{\sin\theta} = \sqrt{5}\), so \(\cos\theta = \sqrt{5}\sin\theta\).
Step 2: Apply the Pythagorean identity.
We know \(\sin^2\theta + \cos^2\theta = 1\). Substituting \(\cos\theta = \sqrt{5}\sin\theta\): \(\sin^2\theta + 5\sin^2\theta = 1\), giving \(6\sin^2\theta = 1\).
Step 3: Solve for \(\sin\theta\).
\(\sin^2\theta = \frac{1}{6}\), so \(\sin\theta = \frac{1}{\sqrt{6}}\) (taking positive value for acute angle).
Step 4: Alternative triangle method.
In a right triangle with opposite \(= 1\) and adjacent \(= \sqrt{5}\), hypotenuse \(= \sqrt{1 + 5} = \sqrt{6}\). So \(\sin\theta = \frac{1}{\sqrt{6}}\).
Step 5: Verify the answer matches option 1.
\(\sin\theta = \frac{1}{\sqrt{6}}\) corresponds to option 1.
Step 6: Eliminate other options.
\(\sqrt{6} > 1\) (invalid for sine), \(\frac{\sqrt{5}}{6}\) is not \(\frac{1}{\sqrt{6}}\), and \(\frac{1}{2}\) would require \(\theta = 30^\circ\) with \(\cot 30^\circ = \sqrt{3} \neq \sqrt{5}\).
\[ \boxed{\dfrac{1}{\sqrt{6}}} \]