PrecalculusHard SharePrecalculus · Hard problemSimplify tanx−sinxsin3x\dfrac{\tan x-\sin x}{\sin^3 x}sin3xtanx−sinx as far as possible.Asecx2+cosx\dfrac{\sec x}{2+\cos x}2+cosxsecxBsecx1+cosx\dfrac{\sec x}{1+\cos x}1+cosxsecxC−secx1+cosx-\dfrac{\sec x}{1+\cos x}−1+cosxsecxD1+cosxsecx\dfrac{1+\cos x}{\sec x}secx1+cosxCheck Show answer Walk through it step by step Practice more like this