Question

In a triangle \( ABC \), \( \angle C = 90^\circ \) and \( \tan A = \dfrac{1}{\sqrt{3}} \). The value of \( \sin A \cos B + \cos A \sin B \) will be:

• A. 0
• B. \( \dfrac{1}{\sqrt{2}} \)
• C. 1
• D. \( \sqrt{2} \)

Hint:

When the sum of two angles equals 90°, use the identity \( \sin A \cos B + \cos A \sin B = \sin (A + B) \) directly to simplify.

Answer 1

The Correct Option is C

Step 1: Use trigonometric identities.
We know that \[ \sin A \cos B + \cos A \sin B = \sin (A + B) \]
Step 2: Relation in a right triangle.
In a right-angled triangle \( ABC \) with \( \angle C = 90^\circ \): \[ A + B = 90^\circ \]
Step 3: Substitute.
\[ \sin (A + B) = \sin 90^\circ = 1 \]
Step 4: Final answer.
\[ \boxed{1} \]