If P + Q + R = 60°, then what is the value of cosQcosR(cosP - sinP) + sinQsinR(sinP - cosP)?
A. $$\frac{1}{2}$$
B. $$\frac{{\sqrt 3 }}{2}$$
C. $$\frac{1}{{\sqrt 2 }}$$
D. $${\sqrt 2 }$$
Answer: Option A
Solution (By Examveda Team)
∵ P + Q + R = 60°By putting P = 0°, Q = 0° and R = 60°
⇒ cosQcosR(cosP - sinP) + sinQsinR(sinP - cosP)
⇒ 1 × cos60°(cos0° - sin0°) + sin0°.sin60°(sin0° - cos0°)
⇒ $$\frac{1}{2}$$ (1 - 0) + 0
⇒ $$\frac{1}{2}$$
This Question Belongs to Arithmetic Ability >> Trigonometry
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