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# Can I multiply both sides of an equation by sec^-1 to get rid of sec on one side?

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Can I multiply both sides of an equation by sec^-1 to get rid of sec on one side?

Here is the problem: sec(Tan^-1(x))=-2/sqrt3

Jan 3, 2016
edited by Guest  Jan 3, 2016
edited by Guest  Jan 3, 2016
edited by Guest  Jan 3, 2016
edited by Guest  Jan 3, 2016

#5
+5

Melody,

arctan x is a multivalued function, it has an infinite number of branches both above and below the one shown by desmos, desmos just shows the principal range.

Care must be taken when giving the values for x.

sec(theta) is negative so theta is in the second or third quadrant, so

$$\displaystyle \tan^{-1}x= \pi \pm \pi/6$$ ,

meaning that

$$\displaystyle x = \tan(\pi \pm \pi/6)$$, angle $$\displaystyle \pm2k\pi$$ for the general solution.

It's tempting to now say that $$\displaystyle x = \pm1/\sqrt{3}$$, but this would be wrong, (without some qualification).

$$\displaystyle x = 1/\sqrt{3}$$, for example, includes the possibility that the angle is $$\displaystyle \pi/6$$, which is not a solution of the original equation.

-Bertie

Jan 5, 2016

#1
+5

sec( arctan(x)) = -2/sqrt(3)

Let arctan(x)  = theta      .....then we have

sec(theta) = -2/sqrt(3)   using the secant inverse

arcsec (-2/sqrt(3))  =  theta = 5pi/6

Then.......we are looking for the tangent of 5pi/6  = x =  -1/sqrt(3)

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As an edit.....the secant could also = -2/sqrt(3) at 7pi/6......in that case, the tangent of 7pi/6 = x  =  1/sqrt(3)   Jan 3, 2016
edited by CPhill  Jan 4, 2016
#2
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Alan or Bertie can you please talk me through this.  I can't bet my head around it :(

sec(Tan^-1(x))= -2/sqrt3

let tan^-1x=theta

then

tan (theta)=x $$sec(\theta) = \sqrt{1+x^2}$$

now it doesn't matter if x is positive or negative sec(theta) has to be positive. SO

I don't think that there should be any solutions.

Here is a graph to back up what I am saying

https://www.desmos.com/calculator/dutfolqi6o

http://www.wolframalpha.com/input/?i=sec%28Tan%5E-1%28x%29%29

BUT

Look at this weird output

http://www.wolframalpha.com/input/?i=sec%28Tan%5E-1%28x%29%29%3D-2%2Fsqrt3

and this one

http://www.wolframalpha.com/input/?i=sec%28Tan%5E-1%28x%29%29%2B2%2Fsqrt3%3D0

I am so confused :((

Jan 4, 2016
#3
+5

Consider, Melody........

Since the secant is negative,  this angle has to be either in the second or third quadrant

If x = -1/sqrt(3)   then    arctan( -1/sqrt(3))  =   -pi/6  = -30°

But this would actually be a 30° angle in the second quadrant since both the secant and arctan are negative.......thus ....arctan (-1/sqrt(3))  = 5pi/6

If x = 1/sqrt(3), then the arctan (1/sqrt(3)) = pi/6  = 30°

But.......this would have to be a 30° angle in the third quadrant because the arctangent of an angle in that quadrant would be positive while, again, the secant is negative, so,  arctan(1/sqrt(3)) = 7pi/6

Check for yourself that  sec (5pi/6)  = sec(7pi/6)  = -2/sqrt(3)   Jan 4, 2016
#4
+5

The problem Chris is

$$\boxed{\frac{-\pi}{2}<tan^{-1}x<\frac{\pi}{2}}$$

You are saying that  tan^-1x = 5pi/6  or  7pi/6

Both these values are outside the range of the function tan^-1x

https://www.desmos.com/calculator/fbnb6rh6md

Jan 4, 2016
edited by Melody  Jan 4, 2016
#5
+5

Melody,

arctan x is a multivalued function, it has an infinite number of branches both above and below the one shown by desmos, desmos just shows the principal range.

Care must be taken when giving the values for x.

sec(theta) is negative so theta is in the second or third quadrant, so

$$\displaystyle \tan^{-1}x= \pi \pm \pi/6$$ ,

meaning that

$$\displaystyle x = \tan(\pi \pm \pi/6)$$, angle $$\displaystyle \pm2k\pi$$ for the general solution.

It's tempting to now say that $$\displaystyle x = \pm1/\sqrt{3}$$, but this would be wrong, (without some qualification).

$$\displaystyle x = 1/\sqrt{3}$$, for example, includes the possibility that the angle is $$\displaystyle \pi/6$$, which is not a solution of the original equation.

-Bertie

Guest Jan 5, 2016
#6
0

ok, thanks Bertie,

It was always drummed into me that the priincipal range, as you call it, was the only one that you were EVER allowed to use.

I can see what you are saying though ://

Jan 5, 2016