# trig identities

how do i solve

secx + tanx = 1

#### jsbeckton

##### Junior Member
break everything down to sines and cosines, look fir identities

#### daon

##### Senior Member
I would do the following:

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.

#### soroban

##### Elite Member
Hello, farfar!

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

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.