I have a problem with understanding why my first method of finding trigonometric function is incorrect.

Question: If $\tan t = \frac{1}{4}$ and the terminal point for $t$ is in Quadrant III, find $\sec t + \cot t$.

Book answer: According to the book I'm using, the answer is: $\frac{16 - \sqrt{17}}{4}$

Solution 1: Here's how I tried to solve the problem:

If $\tan t = \frac{\sin t}{\cos t}$ and it's Quadrant III, then $\sin t = -1$ and $\cos t = -4$

If $\tan t = \frac{1}{4}$, then $\cot t$ is $4$

If $\sec t = \frac{1}{\cos t}$, then $\sec t = -\frac{1}{4}$

So, according to my calculations, $\sec t + \cot t = -\frac{1}{4} + 4 = \frac{15}{4}$, but this doesn't seem to be correct.

I also tried to solve the problem using pythagorean identities and I got correct solution:

If, $\tan^2 t + 1 = \sec^2 t$

$\left(\frac{1}{4}\right)^2 + 1 = \sec^2 t\cdot\frac{1}{16} + 1 = \sec^2 t\cdot \frac{17}{16} = \sec^2 t\cdot \frac{\sqrt{17}}{4} = \sec t$

as it's Quadrant III, $\sec t = - \frac{\sqrt{17}}{4}$

$-\frac{\sqrt{17}}{4} + \frac{16}{4} = \frac{16 - \sqrt{17}}{4}$

I don't understand where things went wrong in Solution 1

  • $\begingroup$ Write $$\cos x=-\frac{1}{\sqrt{1+\tan^2x}}$$ $\endgroup$ – Nosrati Feb 14 '17 at 11:46

$cos x$ can only get values [-1,1] so $cos t$ definitely isn't -4

  • $\begingroup$ You can elaborate your answer and show the correct values of $\cos t$, $\sec t$, etc. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 14 '17 at 11:56
  • $\begingroup$ Thanks for the clue. Indeed range is [-1, 1]. That's the clue why getting sin and cos value from tan is not good idea. Thanks $\endgroup$ – richard.pi Feb 14 '17 at 19:19
  • $\begingroup$ @richard.pi If you want to do it that way you can get them by drawing a triangle with legs $1$ and $4$ and hypotenuse $sqrt(17)$and then calculating $sin$ and $cos$ from that. $\endgroup$ – Abdullah al allah Feb 14 '17 at 20:45

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