Question Number 53328 by Tawa1 last updated on 20/Jan/19 | ||
Commented by maxmathsup by imad last updated on 20/Jan/19 | ||
$${f}\left({x}\right)={tanxsinx}\:=\frac{{sin}^{\mathrm{2}} {x}}{{cosx}}\:. \\ $$ | ||
Commented by Tawa1 last updated on 20/Jan/19 | ||
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19 | ||
$${f}\left({x}\right)={tan}\left(\pi+{x}\right){cos}\left({x}−\frac{\pi}{\mathrm{2}}\right) \\ $$$$=\left({tanx}\right)×{sinx}\: \\ $$$$\:\: \\ $$$$=\frac{{sin}^{\mathrm{2}} {x}}{{cosx}} \\ $$ | ||
Commented by Tawa1 last updated on 20/Jan/19 | ||
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$ | ||