I have trouble proving this. Thanks for any help in advanced!
Prove the following:
tanxsinx tanx - sinx
tanx + sinx = tanxsinx
$\displaystyle \frac{\tan (x)\sin (x)}{\tan (x)+\sin (x)}=\frac{\tan (x)\sin (x)\left( \tan (x)-\sin (x) \right)}{\tan ^{2}(x)\sin ^{2}(x)}=\frac{\tan (x)-\sin (x)}{\tan (x)\sin (x)}.$
Observe that $\displaystyle \tan ^{2}(x)-\sin ^{2}(x)=\sin ^{2}(x)\left( \frac{1}{\cos ^{2}(x)}-1 \right)=\tan ^{2}(x)\sin ^{2}(x),$ and we're done. $\displaystyle \blacksquare$