Question

tana=a/b, then bcos2a+asin2a=?

Answer 1

Answer:

Step-by-step explanation:

Given Tan a=a/b

Sin a/cos a=a/b

Then sin a=a

Cos a=b

We know sin2a=2sina.cosa

Cos2a=2cos^2 a-1

bcos2a+asin2a

b(2cos^2 a -1)+2asina.cosa

2bcos^2 a-b+2asina.cosa

2b(b)^2-b+2a(a)(b)

2b^3-b+2a^2b

b(2a^2+2b^2-1)