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sec(x) + tan(x) = 1

sec(x) = 1 - tan(x)

[sec(x)]^2 = [1 - tan(x)]^2

[sec(x)]^2 = 1 - 2tan(x) + [tan(x)]^2

[sec(x)]^2 - [tan(x)]^2 = 1 - 2tan(x)

But from the identity tan<sup>2</sup>x + 1 = sex<sup>2</sup>x, the left side reduces to 1, thus

1 = 1 - 2tan(x)

0 = -2tan(x)

tan(x) = 0

x = tan<sup>-1</sup>0

Tangent is 0 whenever sine is 0, which is at zero and pi.

I changed to sines and cosines just to see how it turns out . . .Solve: \(\displaystyle \,\sec x\,+\,\tan x\:=\:1\)

We have: \(\displaystyle \L\,\frac{1}{\cos x}\,+\,\frac{\sin x}{\cos x}\;=\;1\)

\(\displaystyle \;\;\)Note that: \(\displaystyle \,\cos x\,\neq\,0\;\;\Rightarrow\;\;x\,\neq\,\frac{\pi}{2},\;\frac{3\pi}{2}\)

Multiply through by \(\displaystyle cos x:\;\;1\,+\,\sin x\;=\;\cos x\)

Square both sides: \(\displaystyle \,1\,+\,2\sin x\,+\,\sin^2x\;=\;\cos^2x\)

Then we have: \(\displaystyle \,1\,+\,2\sin x \,+\,\sin^2x\;=\;1\,-\,\sin^2x\;\;\Rightarrow\;\;2\sin^2x\,+\,2\sin x\;=\;0\)

\(\displaystyle \;\;\;2\sin x(\sin x + 1)\;=\;0\;\;\Rightarrow\;\;\begin{Bmatrix}2\sin x\,=\,0\;\;\Rightarrow\;\;x\,=\,0,\,\pi \\ \sout{\sin x\,=\,-1\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{2},\;\frac{3\pi}{2}}\end{Bmatrix}\)

But the second set of solutions is not in the domain.