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+((1)/(5))^2=1`
`\begin{gathered} \sin ^2\theta=1-(1)/(25) \\ \sin ^2\theta=(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}}{(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}`