Question

Given that $\sin \theta = \frac{a}{b}$, then $\cos \theta$ is equal to :

• A. $\frac{b}{\sqrt{b^2 - a^2}}$
• B. $\frac{b}{a}$
• C. $\frac{\sqrt{b^2 - a^2}}{b}$
• D. $\frac{a}{\sqrt{b^2 - a^2}}$

Hint:

Using a right triangle is often faster:
Draw a triangle with perpendicular = $a$ and hypotenuse = $b$.
The base is $\sqrt{b^2 - a^2}$.
Since $\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}}$, we immediately get $\cos \theta = \frac{\sqrt{b^2 - a^2}}{b}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The topic is Trigonometric Identities and Ratios.
We are given the value of $\sin \theta$ in terms of variables $a$ and $b$ and need to find the expression for $\cos \theta$.

Step 2: Key Formula or Approach:
We can use the fundamental trigonometric identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Rearranging this identity to solve for $\cos \theta$:
\[ \cos^2 \theta = 1 - \sin^2 \theta \]
Taking the square root (assuming $\theta$ is in the first quadrant where trigonometric ratios are positive):
\[ \cos \theta = \sqrt{1 - \sin^2 \theta} \]
Alternatively, we can use a right-angled triangle where:
$\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{a}{b}$
By Pythagoras' Theorem:
$\text{Base} = \sqrt{\text{Hypotenuse}^2 - \text{Perpendicular}^2}$

Step 3: Detailed Explanation:

• Let us use the algebraic approach with the trigonometric identity.
Given:
\[ \sin \theta = \frac{a}{b} \]

• Square both sides:
\[ \sin^2 \theta = \left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2} \]

• Substitute into the rewritten identity:
\[ \cos \theta = \sqrt{1 - \frac{a^2}{b^2}} \]

• Combine terms under the square root by taking a common denominator:
\[ \cos \theta = \sqrt{\frac{b^2 - a^2}{b^2}} \]

• Simplify the fraction by taking the square root of the denominator:
\[ \cos \theta = \frac{\sqrt{b^2 - a^2}}{\sqrt{b^2}} \]
\[ \cos \theta = \frac{\sqrt{b^2 - a^2}}{b} \]


Step 4: Final Answer:
The expression for $\cos \theta$ is $\frac{\sqrt{b^2 - a^2}}{b}$, which matches option (C).