# How do you find the five remaining trigonometric function satisfying sectheta=6/5, tantheta<0?

Jun 20, 2017

$\cos \theta = \frac{5}{6}$; $\sin \theta = - \frac{\sqrt{11}}{6}$; $\tan \theta = - \frac{\sqrt{11}}{5}$; $\cot \theta = - \frac{5}{\sqrt{11}}$; $\csc \theta = - \frac{6}{\sqrt{11}}$

#### Explanation:

Since $\sec \theta = \frac{1}{\cos} \theta = \frac{6}{5}$, you get

$\cos \theta = \frac{1}{\sec} \theta = \frac{5}{6}$

Since

$\sin \theta = \pm \sqrt{1 - {\cos}^{2} \theta}$,

$\tan \theta < 0$,

$\cos \theta > 0$,

$\tan \theta = \sin \frac{\theta}{\cos} \theta$,

then

$\sin \theta = - \sqrt{1 - {\cos}^{2} \theta} = - \sqrt{1 - \frac{25}{36}} = - \frac{\sqrt{11}}{6}$

and

$\tan \theta = \frac{- \frac{\sqrt{11}}{\cancel{6}}}{\frac{5}{\cancel{6}}} = - \frac{\sqrt{11}}{5}$

$\cot \theta = \frac{1}{\tan} \theta = - \frac{5}{\sqrt{11}}$

$\csc \theta = \frac{1}{\sin} \theta = - \frac{6}{\sqrt{11}}$

Jun 20, 2017

$\sin \theta \text{ "cos theta" "tan theta" "cot theta" "sec theta" cosec "theta}$

$\frac{- \sqrt{11}}{6} \text{ "5/6" "(-sqrt11)/5" "5/(-sqrt11)" "6/5" } \frac{6}{- \sqrt{11}}$

#### Explanation:

$\sec \theta = \frac{6}{5} = \frac{r}{x}$

Therefore we can find $y$ using Pythagoras' Theorem.

$y = \sqrt{{6}^{2} - {5}^{2}} = \sqrt{11}$

We need to decide which quadrant we are in.

$\sec \theta$ is given as positive $\rightarrow 1 s t \mathmr{and} 4 t h$
$\tan \theta$ is given as negative $\rightarrow 2 n d \mathmr{and} 4 t h$

So the only quadrant where both condition hold is the $4 t h$

In the $4 t h$ quadrant, only $\cos \theta \mathmr{and} \sec \theta$ are positive.

We can now write all $6$ ratios:

$x = 5 , \text{ "y =-sqrt11" " and " } r = 6$

$\sin \theta \text{ "cos theta" "tan theta" "cot theta" "sec theta" cosec "theta}$

$- \frac{y}{r} \text{ "x/r" "-y/x" "-x/y" "r/x" "-r/y}$

$\frac{- \sqrt{11}}{6} \text{ "5/6" "(-sqrt11)/5" "5/(-sqrt11)" "6/5" } \frac{6}{- \sqrt{11}}$