Question

If $\frac{{\tan }^3\alpha }{1+{\tan }^2\alpha }=\sec \alpha \sin \alpha -a\sin \alpha \cos \alpha$. Find $a$

Answer 1

LHS $\frac{{\tan }^3\alpha }{1+{\tan }^2\alpha }$

$=\frac{\frac{{\sin }^3\alpha }{{\cos }^3\alpha }}{1+\frac{{\sin }^2\alpha }{{\cos }^2\alpha }}=\frac{\frac{{\sin }^3\alpha }{{\cos }^3\alpha }}{\frac{{\cos }^2\alpha +{\sin }^2\alpha }{{\cos }^2\alpha }}$

$\because {\sin }^{2}x+{\cos }^{2}x=1$

$=\frac{{\sin }^3\alpha {\cos }^2\alpha }{{\cos }^3\alpha }=\frac{{\sin }^3\alpha }{\cos \alpha }$

$=\frac{\sin \alpha \left({\sin }^2\alpha \right)}{\cos \alpha }$

$=\frac{\sin \alpha (1-{\cos }^2\alpha )}{\cos \alpha }$

$=\frac{\sin \alpha -\sin \alpha {\cos }^2\alpha }{\cos \alpha }$

$=\frac{\sin \alpha }{\cos \alpha }-\sin \alpha \cos \alpha$

$=\sin \alpha \sec \alpha -\sin \alpha \cos \alpha$

$\Rightarrow \alpha =1$