Please help me with the answer

35. If P + Q + R = 60°, then what is the value of cos Q cos R (cos P – sin P) + sin Q sin R (sin P – cos P)
(1) $\frac{1}{2}$
(2) $\frac{\sqrt{3}}{2}$
(3) $\frac{1}{\sqrt{2}}$
(4) $\sqrt{2}$

Dear Student
ATQ,
P + Q + R = 60°
Let P = 0°, Q = 0° => R = 60°
cosQ. cosR (cosP – sinP) + sinQ sinR (Sin P – CosP)
= cos0°. cos60° (cos0° – sin0°) + sin 0°. sin 60° (sin0° – cos0°)
 = 1. 1/ 2 (1 – 0) + 0 × sin 60°. (sin 0 – cos0°)
= 1 /2 + 0 = 1 /2
Option A
Regards