Using the definations of sec, cosec, cot and tan simplify the following expression;
$\displaystyle sec^2x cos^5x + cotx \ cosecx \ sin^4x $
we have
$\displaystyle secx=\frac{1}{cosx}$
$\displaystyle cotx=\frac{cosx}{sinx}$
$\displaystyle cosecx=\frac{1}{sinx}$
$\displaystyle =(\frac{1}{cosx})^2.\ cos^5x + \frac{cosx}{sinx} .\ \frac{1}{sinx}. \ sin^4x $
$\displaystyle =cos^3x+cosx\sin^2x$
$\displaystyle =cosx(cos^2x+sin^2x)$
$\displaystyle =cosx$.............ANSWER
since $\displaystyle cos^2x+sin^2x=1$