Question

In \( \triangle PQR \), \( \angle Q = 90^\circ \), if \( \tan P = \frac{1}{\sqrt{3}} \), then find the value of \( \sin P \cos R + \cos P \sin R \).

Hint:

Use the identity \( \sin(P + R) = \sin P \cos R + \cos P \sin R \) when \( P + R = 90^\circ \) in a right triangle.

Answer 1

We are given that \( \angle Q = 90^\circ \), so \( \triangle PQR \) is a right triangle. Also, we are given that \( \tan P = \frac{1}{\sqrt{3}} \). We need to find the value of \( \sin P \cos R + \cos P \sin R \). Since \( \angle P + \angle R = 90^\circ \) (because the sum of angles in a triangle is \( 180^\circ \), and \( \angle Q = 90^\circ \)), we know that: \[ \angle R = 90^\circ - \angle P. \] We can use the trigonometric identity: \[ \sin(P + R) = \sin P \cos R + \cos P \sin R. \] Since \( P + R = 90^\circ \), we have: \[ \sin(P + R) = \sin 90^\circ = 1. \] Thus, \[ \sin P \cos R + \cos P \sin R = 1. \]
Conclusion:
The value of \( \sin P \cos R + \cos P \sin R \) is \( 1 \).