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 trigonometric identities:
\( \csc^2 \theta = 1 + \cot^2 \theta = 1 + 9 = 10 \implies \sin\theta = \frac{1}{\sqrt{10}} \).
Since \( \cos\theta = \cot\theta \cdot \sin\theta \), we get \( \cos\theta = 3 \cdot \frac{1}{\sqrt{10}} = \frac{3}{\sqrt{10}} \).
Identity methods are often faster than drawing triangles!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question provides the value of \(\cot \theta\) and asks us to find the value of \(\cos \theta\).
This requires applying trigonometric ratios and right-angled triangle properties.

Step 2: Key Formula or Approach:
In a right-angled triangle, the trigonometric ratios are defined as:
\[ \cot \theta = \frac{\text{Adjacent Side}}{\text{Opposite Side}} = \frac{\text{Base}}{\text{Perpendicular}} \]
\[ \cos \theta = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} = \frac{\text{Base}}{\text{Hypotenuse}} \]
We can use Pythagoras' theorem to find the hypotenuse:
\[ \text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2 \]

Step 3: Detailed Explanation:
1. We are given \( \cot \theta = 3 \).
2. We can write this ratio as \( \cot \theta = \frac{3}{1} \).
3. Let the Base (\( B \)) be \( 3k \) and the Perpendicular (\( P \)) be \( 1k \), where \( k \) is a positive constant.
4. According to Pythagoras' theorem, the Hypotenuse (\( H \)) is:
\[ H = \sqrt{P^2 + B^2} \]
\[ H = \sqrt{(1k)^2 + (3k)^2} = \sqrt{k^2 + 9k^2} = \sqrt{10k^2} = k\sqrt{10} \]
5. Now, we calculate the value of \( \cos \theta \):
\[ \cos \theta = \frac{B}{H} = \frac{3k}{k\sqrt{10}} = \frac{3}{\sqrt{10}} \]
6. This matches option (C).

Step 4: Final Answer:
The correct option is (C).