trig identities

farfar

New member
Joined
Apr 26, 2006
Messages
1
how do i solve

secx + tanx = 1
 

jsbeckton

Junior Member
Joined
Oct 24, 2005
Messages
174
break everything down to sines and cosines, look fir identities
 

daon

Senior Member
Joined
Jan 27, 2006
Messages
1,284
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
Joined
Jan 28, 2005
Messages
5,588
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.
 
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