The question is :

Let $\cos\theta= \frac{-1}{4}$ and $\tan \theta> 0$. Find the remaining trigonometric functions.

I got

$\sin \theta = -\frac {\sqrt{15}}{4}$

$\tan \theta = \sqrt {15}$

$\csc \theta= \frac{-4\sqrt{15}}{15}$

$\sec \theta= -4$

$\cot \theta= \frac{\sqrt{15}}{15}$

Is this correct?

  • $\begingroup$ That many questions asked and still such poor formatting. It's high time you learn MathJax. $\endgroup$ – Shraddheya Shendre Feb 6 '17 at 8:45
  • $\begingroup$ I think all these correct. $\endgroup$ – Nosrati Feb 6 '17 at 8:48
  • $\begingroup$ Sine must be negative...which fits what you wrote about cosecant... $\endgroup$ – DonAntonio Feb 6 '17 at 9:30
  • $\begingroup$ @ShraddheyaShendre Please do pay attention when you edit: you ommited a minus sign in $\;\sin\theta\;$ ... $\endgroup$ – DonAntonio Feb 6 '17 at 10:00
  • $\begingroup$ @DonAntonio - I am not sure whether it was a minus sign or a hyphen. But seeing the negative cosec, I now think that it indeed was a minus sign. Sorry. $\endgroup$ – Shraddheya Shendre Feb 6 '17 at 15:15

No. They all are correct except $Sin\hspace{1mm}\theta$.
From the information given in question, we are certain that the quadrant talked about is 3rd. Since $Tan\hspace{1mm}\theta>0$ and $Cos\hspace{1mm}\theta=-\frac{1}{4}$
So we can use $Sin^2\hspace{1mm}\theta+Cos^2\hspace{1mm}\theta=1$ and $Sin\hspace{1mm}\theta<0$ in 3rd quadrant.
So $Sin\hspace{1mm}\theta$ is negative.
And then, everything else is straightforward. I hope this helps.

  • $\begingroup$ Then no: not all are correct since, as you say, we're in the third quadrant and thus sine must be negative ... $\endgroup$ – DonAntonio Feb 6 '17 at 9:29
  • $\begingroup$ Wow. I don't know how I didn't notice that before. Yes, sine will be negative. Thanks for pointing that out! $\endgroup$ – Sum-Meister Feb 6 '17 at 9:34
  • $\begingroup$ @DonAntonio In main post there was a negative back of sine, in edit it changed. $\endgroup$ – Nosrati Feb 6 '17 at 9:41
  • $\begingroup$ @MyGlasses True indeed, I just checked. $\endgroup$ – DonAntonio Feb 6 '17 at 10:00

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