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trig identities
- Thread starter farfar
- Start date
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.
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.
Hello, farfar!
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.
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.