Question

If $tan\theta =\frac{a}{b}$, prove that $asin2\theta +bcos2\theta =b$

Answer 1

$asin2\theta +bcos2\theta =a\times \frac{2ta{n}_2\theta }{1+ta{n}^2\theta }+b\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }=a\times \frac{\frac{2a}{b}}{1+\frac{a^2}{b^2}}+b\frac{1-\frac{a^2}{b^2}}{1+\frac{a^2}{b^2}}=\frac{\frac{2a^2}{b}}{\frac{b^2+a^2}{b^2}}+\frac{b\left(\frac{b^2-a^2}{b^2}\right)}{\frac{b^2+a^2}{b^2}}+\frac{b^2+a^2}{b^2}=\frac{2a^{2}b}{b^2+a^2}+\frac{b^2{a}^{2}b}{b^2+a^2}=\frac{2a^{2}b+b^3-a^{2}b}{b^2+a^2}=\frac{a^{2}b+b^3}{a^2+b^2}=\frac{b(a^2+b^2)}{a^2+b^2}=b(=RHS)$