If $y=\tan x.{\cos }^{2}x$ then $\frac{dy}{dx}$ will be

The correct option is B $1-2{\sin }^{2}x$
$y=tanxco{s}^{2}x\frac{dy}{dx}=tanx\frac{d}{dx}\left(co{s}^{2}x\right)+co{s}^{2}x\frac{d}{dx}\left(tanx\right)=tanx\left(2cosx\right)(-sinx)+co{s}^{2}xse{c}^{2}x=1-2si{n}^{2}x$