If `tantheta=(a)/(b)`, then `bcos2theta+asin2theta` is equal to

Options

A. `a`
B. `b `
C. `(a)/(b)`
D. None of these

Answer

B

Explanation

Given that, `tantheta=(a)/(b)`
`therefore" "bcos2theta+asin2theta=b((1-tan^(2)theta)/(1+tan^(2)theta))+a((2tantheta)/(1+tan^(2)theta))`
`" "=b((1-(a^(2))/(b^(2)))/(1+(a^(2))/(b^(2))))+a(((2a)/(b))/(1+(a^(2))/(b^(2))))`
`" "=b((b^(2)-a^(2))/(b^(2)+a^(2)))+(2a^(2)b)/(a^(2)+b^(2))`
`" "=(b)/(a^(2)+b^(2))[b^(2)-a^(2)+2a^(2)]=((a^(2)+b^(2))b)/((a^(2)+b^(2)))`
=b