Question

In a triangle ABC, $\angle C = 90^\circ$ and $\tan A = \frac{1}{\sqrt{3}}$. The value of the following will be: \[ \sin A \cos B + \cos A \sin B \]

• A. 0
• B. $\frac{1}{\sqrt{2}}$
• C. 1
• D. $\sqrt{2}$

Hint:

The sine addition formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$ is useful for finding the value of expressions involving angles of a triangle.

Answer 1

The Correct Option is C

We are given that $\tan A = \frac{1}{\sqrt{3}}$. This implies: \[ \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}} \Rightarrow \text{opposite} = 1, \quad \text{adjacent} = \sqrt{3}. \] Using the Pythagorean theorem, we can find the hypotenuse: \[ \text{hypotenuse} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2. \] Thus, in triangle ABC: \[ \sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{2}, \quad \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2}. \] Next, we need to find $\sin A \cos B + \cos A \sin B$. By the sine addition formula: \[ \sin(A + B) = \sin A \cos B + \cos A \sin B \] Since $\angle C = 90^\circ$, we know that: \[ A + B = 90^\circ \Rightarrow \sin(A + B) = \sin 90^\circ = 1. \] Thus, we have: \[ \sin A \cos B + \cos A \sin B = 1. \]
Step 2: Conclusion.
Therefore, the value of the expression is $1$. So, the correct answer is (C).