Question

Hi, I was absent today in class and missed the whole lesson, this is not a grade exam this is a homework that my teacher assigned us to do, can you please explain how to do it much appreciated it

Answer 1

we have

sec(θ)=5 and tan(θ) > 0

that means

The angle θ lies on the I quadrant

step 1

Find out the value of cosθ

Remember that

secθ=1/cosθ

5=1/cosθ

cosθ=1/5

step 2

Find out the value of sinθ

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

substitute the value of the cosine

$\sin ^2\theta+(\frac{1}{5})^2=1$$\begin{gathered} \sin ^2\theta=1-\frac{1}{25} \\ \sin ^2\theta=\frac{24}{25} \end{gathered}$$\sin ^{}\theta=\pm\frac{2\sqrt[]{6}}{5}$

Remember that the angle lies on the first quadrant

so

the value of the sine is positive

therefore

$\sin ^{}\theta=\frac{2\sqrt[]{6}}{5}$

step 3

Find out the value of the tanθ

tanθ=sinθ/cosθ

substitute given values

$\tan \theta=\frac{\frac{2\sqrt[]{6}}{5}}{\frac{1}{5}}=2\sqrt[]{6}$

step 4

Find out the value of cotθ

cotθ=1/tanθ

substitute the value of tanθ

$\cot \theta=\frac{1}{2\sqrt[]{6}}$

simplify

$\cot \theta=\frac{1}{2\sqrt[]{6}}\cdot\frac{\sqrt[]{6}}{\sqrt[]{6}}=\frac{\sqrt[]{6}}{12}$

step 5

Find out the value of cscθ

cscθ=1/sinθ

substitute the value of the sine

$\csc \theta=\frac{1}{\frac{2\sqrt[]{6}}{5}}=\frac{5}{2\sqrt[]{6}}$

simplify

$\csc \theta=\frac{5}{2\sqrt[]{6}}\cdot\frac{\sqrt[]{6}}{\sqrt[]{6}}=\frac{5\sqrt[]{6}}{12}$