Question

Given cot \(\theta\) = 3, the value of cos \(\theta\) is :

• A. \(\frac{1}{3}\)
• B. \(\frac{1}{\sqrt{10}}\)
• C. \(\frac{3}{\sqrt{10}}\)
• D. \(\frac{\sqrt{10}}{3}\)

Hint:

Alternatively, use the identity \(\text{cosec}^2 \theta = 1 + \cot^2 \theta\).
\(\text{cosec}^2 \theta = 1 + 3^2 = 10 \implies \sin \theta = \frac{1}{\sqrt{10}}\).
Then, \(\cos \theta = \cot \theta \times \sin \theta = 3 \times \frac{1}{\sqrt{10}} = \frac{3}{\sqrt{10}}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In a right-angled triangle, the cotangent ratio is the ratio of the adjacent side to the opposite side. The cosine ratio is the ratio of the adjacent side to the hypotenuse.
Step 2: Key Formula or Approach:
1. \(\cot \theta = \frac{\text{Base (B)}}{\text{Perpendicular (P)}}\)
2. Pythagoras Theorem: \(H = \sqrt{P^2 + B^2}\)
3. \(\cos \theta = \frac{B}{H}\)
Step 3: Detailed Explanation:
Given \(\cot \theta = 3\), we can write this as \(\frac{3}{1}\).
Let Base (\(B\)) \( = 3k \) and Perpendicular (\(P\)) \( = 1k \).
Calculate Hypotenuse (\(H\)):
\[ H = \sqrt{(3k)^2 + (1k)^2} = \sqrt{9k^2 + k^2} = \sqrt{10k^2} = k\sqrt{10} \]
Now, find \(\cos \theta\):
\[ \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{3k}{k\sqrt{10}} = \frac{3}{\sqrt{10}} \]
Step 4: Final Answer:
The value of \(\cos \theta\) is \(\frac{3}{\sqrt{10}}\).