# Let θ be an angle in quadrant II such that sinθ= (1/4), how do you find the values of secθ and tanθ?

Sep 4, 2016

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

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

#### Explanation:

Recall that $\sin \theta = \text{opposite"/"hypotenuse}$

Hence, the side opposite $\theta$ in our question measures $1$ unit and the hypotenuse measures $4$ units.

Since we're dealing with right triangles, we can find the side adjacent $\theta$ using pythagorean theorem.

Let the adjacent side be $a$.

${a}^{2} + {1}^{2} = {4}^{2}$

${a}^{2} + 1 = 16$

${a}^{2} = 15$

$a = \sqrt{15}$

Now, let's define secant and tangent.

sectheta = 1/(costheta) = 1/("adjacent"/"hypotenuse") = "hypotenuse"/"adjacent"

tantheta = sintheta/costheta = ("opposite"/"hypotenuse")/("adjacent"/"hypotenuse") = "opposite"/"adjacent"

Applying these definitions:

$\sec \theta = \frac{4}{\sqrt{15}} = \frac{4 \sqrt{15}}{15}$

$\tan \theta = \frac{1}{\sqrt{15}} = \frac{\sqrt{15}}{15}$

The last thing left to do is to find the signs of these ratios. We know that we're in quadrant $I I$, where sine is positive, and all the other ratios are negative. Since secant is related to cosine, it will be negative.

So, our final ratios are:

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

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

Hopefully this helps!