Question

Find sec theta if theta is in quadrant 4 and sin theta= -1/5

Answer 1

Answer:

$\sec(\theta)=\frac{5\sqrt{6}}{12}$

The answer is the last one.

Step-by-step explanation:

If we are in quadrant 4, then x (cosine) is positive and y (sine) is negative.

Since cosine is positive, secant is positive because secant is the reciprocal of cosine.

So we already know the answer is not the 1st one or the 3rd one.

I'm going to use a Pythagorean Identity to find cosine value of theta.

$\cos^2(\theta)+\sin^2(\theta)=1$

Enter in -1/5 for $\sin(\theta)$:

$\cos^2(\theta)+(\frac{-1}{5})^2=1$

Simplify a bit:

$\cos^2(\theta)+\frac{1}{25}=1$

Subtract 1/25 on both sides:

$\cos^2(\theta)=1-\frac{1}{25}$

Write 1 as 25/25 so you have a common denominator on the right hand side:

$\cos^2(\theta)=\frac{25}{25}-\frac{1}{25}$

$\cos^2(\theta)=\frac{24}{25}$

Take the square root of both sides:

$\cos(\theta)=\pm \sqrt{\frac{24}{25}}$

$\cos(\theta)=\pm \frac{\sqrt{24}}{\sqrt{25}}$

I will worry about simplifying the square root part when finding secant.

We said that cosine was positive because we were in the fourth quadrant.

$\cos(\theta)=\frac{\sqrt{24}}{\sqrt{25}}$

Now recall that cosine and secant are reciprocals of each other:

$\sec(\theta)=\frac{\sqrt{25}}{\sqrt{24}}$

Let's simplify the square part not.

Usually people hate the square root on both and also if you look at your choices none of the choice have square root on bottom.

So we are going to multiply top and bottom by $\sqrt{24}$. I'm going to also write 5 instead of $\sqrt{25}$.

$\sec(\theta)=\frac{5}{\sqrt{24}} \cdot \frac{\sqrt{24}}{\sqrt{24}}$

$\sec(\theta)=\frac{5\sqrt{24}}{24}$

Now let's simplify the square root of 24.

We know 24 is not a perfect square, but 24 does contain a factor that is a perfect square. That factor is 4.

$\sec(\theta)=\frac{5\sqrt{4}\sqrt{6}}{24}$.

$\sec(\theta)=\frac{5(2)\sqrt{6}}{24}$

$\sec(\theta)=\frac{10\sqrt{6}}{24}$

Now both 10 and 24 share a common factor of 2 so let's divide top and bottom by 2:

$\sec(\theta)=\frac{5\sqrt{6}}{12}$

The answer is the last one.

Answer 2

Answer:

$\frac{5}{2\sqrt{6} }$

Step-by-step explanation:

Since Θ is in fourth quadrant then cosΘ > 0, as is secΘ

Given

sinΘ = - $\frac{1}{5}$, then

cosΘ = $\sqrt{1-(-1/5)^2}$

         = $\sqrt{1-\frac{1}{25} }$ = $\sqrt{\frac{24}{25} }$ = $\frac{2\sqrt{6} }{5}$

Hence

secΘ = $\frac{1}{\frac{2\sqrt{6} }{5} }$ = $\frac{5}{2\sqrt{6} }$