© examsiri.com
Question : 21 of 160
Marks:
+1,
-0
Solution:
21. (a) In
△ABC,A+B+C=π⇒A+B=π−C Now,
sin‌2A+sin‌2B+sin‌2C.
‌=2‌sin‌(‌)‌cos(‌)+2‌sin‌C⋅cos‌C ‌=2‌sin‌(A+B)⋅cos(A−B)+2‌sin‌C⋅cos‌C =2‌sin(π−C).cos(A−B)+2‌sin.cos‌C =2‌sin‌C[cos(A−B)+cos‌C] =2‌sin‌C[cos(A−B)+cos(π−A+B)] =2‌sin‌C[cos(A−B)−cos(A+B)] =2‌sin‌C×2‌sin‌A‌Sin‌B=4‌sin‌A.sin‌B.sin‌C
© examsiri.com
Go to Question: