?
If P + Q + R = 60°, then what is the value of cosQcosR(cosP - sinP) + sinQsinR(sinP - cosP)?
Answer & Solution
Correct Answer:
Option
A
∵ 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
⇒ $$\frac{1}{2}$$
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
⇒ $$\frac{1}{2}$$
Join the Discussion
Login to post a comment or share your explanation.
Login