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: Understanding the Question:
We are given the value of \(\cot \theta = \sqrt{5}\) for an angle \(\theta\). We need to determine the value of \(\sin \theta\) using trigonometric ratios or identities.

Step 2: Key Formula or Approach:
We can use a right-angled triangle approach or trigonometric identities.
1. Right-triangle approach: \[ \cot \theta = \frac{\text{Base (B)}}{\text{Perpendicular (P)}} = \frac{\sqrt{5}}{1} \] By Pythagoras theorem: \[ \text{Hypotenuse (H)} = \sqrt{\text{P}^2 + \text{B}^2} \] Then, find the sine ratio: \[ \sin \theta = \frac{\text{Perpendicular (P)}}{\text{Hypotenuse (H)}} \]

Step 3: Detailed Explanation:
1. Let us assume a right-angled triangle where the ratio of base to perpendicular is: \[ \text{Base (B)} = \sqrt{5} \] \[ \text{Perpendicular (P)} = 1 \] 2. Apply the Pythagorean Theorem to find the Hypotenuse (H): \[ H = \sqrt{P^2 + B^2} \] \[ H = \sqrt{(1)^2 + (\sqrt{5})^2} \] \[ H = \sqrt{1 + 5} \] \[ H = \sqrt{6} \] 3. Now, find the value of \(\sin \theta\): \[ \sin \theta = \frac{P}{H} \] \[ \sin \theta = \frac{1}{\sqrt{6}} \]

Step 4: Final Answer:
The value of \(\sin \theta\) is \(\frac{1}{\sqrt{6}}\).
Thus, the correct option is (A).