Question

If $\frac{cosx}{a}=\frac{sinx}{b}$ then $|acos2x+bsin2x|=$

• A. |a|
• B. $|a\sqrt{ab}|$
• C. $\frac{a^2}{|b|}$
• D. $\frac{b^2}{|a|}$

Answer 1

The correct option is A

|a|

$\frac{cosx}{a}=\frac{sinx}{b}$
$\Rightarrow tanx=\frac{b}{a}$
$\sin 2x=\frac{2tanx}{1+ta{n}^{2}x}=\frac{2ab}{a^2+b^2}cos2x=\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}$
$\Rightarrow cos2x=\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}$
$\Rightarrow cos2x=\frac{a^2-b^2}{a^2+b^2}acos2x+bsin2x=\frac{a(a^2-b^2)}{a^2+b^2}+b\frac{\left(2ab\right)}{a^2+b^2}=\frac{a^3-a{b}^2+2a{b}^2}{a^2+b^2}=\frac{a^3+a{b}^2}{a^2+b^2}=\frac{(a^2+b^2)}{a^2+b^2}=a$