please answer this question faster
Attachments:
Answers
Answered by
1
Answer:
Option(B)
Step-by-step explanation:
Given :
A + B + C = 180° Then, B + C = 180° - A
Now,
cos 2A + cos 2B + cos 2C + 1
⇒ cos2A + [cos 2B + cos 2C] + 1
⇒ cos 2A + [2 * cos(B + C) * cos(B - C)] + 1
⇒ (2 cos²A - 1) + [2 cos(180 - A) * cos(B - C)] + 1
⇒ 2 cos²A + [-2 cosA * cos(B - C)]
⇒ 2 cosA[cos(A) - cos(B - C)]
⇒ 2cosA[cos{180 - (B + C)} - cos(B - C)]]
⇒ 2cosA[-cos(B + C) - cos(B - C)]
⇒ -2cosA[cos(B + C) - cos(B - C)]
⇒ -2cosA[2cosBcosC]
⇒ -4cosAcosBcosC
Hope it helps!
Similar questions