f(x)=secxcscx
let secx=g(x)
let cscx=h(x)
therefore, using leibnitz rule,
f'(x)=g'(x)*h(x)+h'(x)*g(x)
Since derivative of secx=secxtanx
and derivative of cscx=-cscxcotx,
f'(x)=secxtanxcscx -cscxcotxsecx
=1/cosx*sinx/cosx*1/sinx - 1/sinx*cosx/sinx*1/cosx
=1/cos^2x-1/sin^2x
=(sin^2x-cos^2x)/sin^2xcos^2x / Option1
