If $\tan x-\sin x=a$, what is $\sec^3(x)-1$ in terms of $a$ I have found that $\sin x\cdot(\sec x-1)=a$ so $\sec x=\frac a{\sin x} +1$. I need help to proceed.
 A: We can write
$$
\tan x-\sin x = \sin x(\sec x-1) 
$$
then
$$
(\tan x-\sin x)^2=\sin^2x(\sec x-1)^2=(1-\cos^2x)(\sec x-1)^2=\frac{(\sec^2x-1)(\sec x-1)^2}{\sec^2x}
$$
So we can try to solve the equation
$$
(\sec^2x-1)(\sec x-1)^2=a^2\sec^2x
$$
or, setting $s = \sec x$,
$$
s^4-2s^3-a^2s^2+2s-1=0
$$
but it doesn't seem to be easy.
A: $$Define\; the\; function\;\;;\;\\~\\f(x)＝\tan(x)－\sin(x)\\~\\f′(x)＝0\;\;\;\;\;\;\;\text{because}\;\;f(x)＝a\\~\\f′(x)＝1＋\tan^2(x)－\cos(x)＝0\\~\\ \tan^2(x)＝\cos(x)－1\\~\\ \tan^2(x)\ge \;0\;\\~\ \cos(x)－1\;\ge 0\;\;\implies\;x＝2\pi\;k\;\;where\;k\in\;Z\\~\ \text{then}\;;\;\sec^3(x)－1＝\sec^3(2\pi\;k)-1＝0\\~\\ \sec^3(x)－1＝0$$
A: We can recall the difference of cubes formula
$$\ a^3 - b^3 = (a - b)(a^2 + 2ab + b^2) $$
So now we see that 
$$\ \sec^3x - 1 =  (\sec x - 1)(\sec^2x - 2\sec x + 1) $$
Or that 
$$\ \sec^3x - 1 = (A/\sin x)(\sec^2x - 2\sec x + 1) $$
Continue with the simplification process. 
$$\ \sec^3x - 1 = A/\sin x(2 - 2\sec x + \tan^2x)$$
$$\ \sec^3x - 1 = A/\sin x(-A/\sin x + \tan^2x) $$
$$\ \sec^3x - 1 = (A\sin x)/\cos^2x - (A^2)/(\sin^2x)$$
We can't go much further than that, unfortunately.  I would like to see if any contributors have other ideas.  
