Answer
a. $2csc^3x-csc(x)$,
b. $2sec^3x-sec(x)$,
Work Step by Step
Recall the formulas $\frac{d}{dx}(tan(x))=sec^2x$, $\frac{d}{dx}(cot(x))=-csc^2x$, $\frac{d}{dx}(sec(x))=sec(x)tan(x)$, $\frac{d}{dx}(csc(x))=-csc(x)cot(x)$,
a. $y'=-csc(x)cot(x)$, $y''=csc(x)csc^2x+csc(x)cot(x)cot(x)=csc^3x+csc(x)cot^2(x)=csc^3x+csc(x)(csc^2(x)-1)=2csc^3x-csc(x)$,
b. $y'=sec(x)tan(x)$, $y''=sec(x)sec^2x+sec(x)tan(x)tan(x)=sec^3x+sec(x)tan^2(x)=sec^3x+sec(x)(sec^2(x)-1)=2sec^3x-sec(x)$,