Question

How do you prove `sec x + sin x = tan x sin x`?

Answer 1

This is true only for `x=-\pi/4` plus multiples of `\pi`.

Explanation:

Since `\sec(x)=1/\cos(x)` and `\tan(x)=\sin(x)/\cos(x)`, first multiply by `\cos(x)`. Thus

`1+\sin(x)\cos(x)=\sin^2(x)`

Put in #1=\sin^2(x)+\cos^2(x):

`\sin^2(x)+\cos^2(x)+\sin(x)\cos(x)=\sin^2(x)`

`\cos^2(x)+\sin(x)\cos(x)=0`

`\cos(x)(\cos(x)+\sin(x))=0`

Thus the equation is true only if `\cos(x)=0` or `\cos(x)=-sin(x)`.

But we cannot accept `\cos(x)=0` because that leaves `\sec(x)=1/\cos(x)` undefined. So `\cos(x)=-sin(x)`, thus `x=-\pi/4` plus multiples of `\pi`.