1.

If `tantheta=(a)/(b)`, then `bcos2theta+asin2theta` is equal toA. aB. bC. `(a)/(b)`D. None of these

Answer» Correct Answer - B
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


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