Question

If $a\tan \theta =b$, then $a\cos \theta +\sin 2\theta =$

Answer 1

$a\tan \theta =b$

$\Rightarrow \tan \theta =\frac{b}{a}$

$\Rightarrow \sin \theta =\frac{b}{\sqrt{a^2+b^2}}$

$\cos \theta =\frac{a}{\sqrt{a^2+b^2}}$

$\Rightarrow \sin 2\theta =\frac{2ab}{a^2+b^2}$

$\Rightarrow a\cos \theta +\sin 2\theta =\frac{a^2}{\sqrt{a^2+b^2}}+\frac{2ab}{{\left(\sqrt{a^2+b^2}\right)}^2}$

$=\frac{a^2}{\sqrt{a^2+b^2}}+\frac{2ab}{a^2+b^2}.$

Hence, solved.