Question

If \(\cos \theta = x\) then \(\tan \theta = \)

• A. \(\frac{\sqrt{1+x^2}}{x}\)
• B. \(\frac{\sqrt{1-x^2}}{x}\)
• C. \(\sqrt{1-x^2}\)
• D. \(\frac{x}{\sqrt{1-x^2}}\)

Hint:

When given one trigonometric ratio and asked for another, drawing a simple right-angled triangle is often the most intuitive and error-free method.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to express \(\tan \theta\) in terms of \(x\), given that \(\cos \theta = x\). We can do this using trigonometric identities or by constructing a right-angled triangle.

Step 2: Key Formula or Approach:
Using identities:
1. The Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\), which gives \(\sin \theta = \sqrt{1 - \cos^2 \theta}\) (assuming \(\theta\) is in the first quadrant). 2. The ratio identity: \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).

Step 3: Detailed Explanation:
We are given \(\cos \theta = x\).
First, find \(\sin \theta\) using the Pythagorean identity:
\[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2} \] Now, use the ratio identity to find \(\tan \theta\):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{1 - x^2}}{x} \] Alternatively, using a triangle:
Since \(\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{x}{1}\), we can set Adjacent = \(x\) and Hypotenuse = 1.
By Pythagorean theorem, Opposite = \(\sqrt{\text{Hypotenuse}^2 - \text{Adjacent}^2} = \sqrt{1^2 - x^2} = \sqrt{1 - x^2}\).
Then, \(\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sqrt{1 - x^2}}{x}\).

Step 4: Final Answer:
The value of \(\tan \theta\) is \(\frac{\sqrt{1-x^2}}{x}\).