Question

help me please. i want to know if im doing it right. if sec (θ) = 5/3 and 0° < θ < 90°, what is cot (θ)?write the answer in simplified rationalized form.

Answer 1

If we are working with angles between 0° and 90°, we can use a right triangle to help us visualize the answer.

So, we know that secant is the inverse of cosine, that is:

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

And we know that cosing, in a right triangle, is the adjancet leg divided by the hypotenuse. Let's call l_a the adjacent leg, l_o the opposide leg and h the hypotenuse. So:

$\sec \theta=\frac{1}{\cos \theta}=\frac{1}{\frac{l_a}{h}}=\frac{h}{l_a}$

And we know that sec (θ) is 5/3, so if we draw a triangle with l_a = 1, we will have:

$h=l_a\sec \theta=1\cdot\frac{5}{3}=\frac{5}{3}$

so:

The cot(θ) is the inverse of tan(θ), so:

$\cot (\theta)=\frac{1}{\tan(\theta)}=\frac{1}{\frac{l_o}{l_a}}=\frac{l_a}{l_o}$

So, we can use the Pythagora's theorem to find l_o and calculate cot(θ):

$\begin{gathered} (\frac{5}{3})^2=1^2+l^2_o_{} \\ l^2_o=\frac{25}{9}-1=\frac{25-9}{9}=\frac{16}{9} \\ l_o=\frac{4}{3} \end{gathered}$

So:

$\cot (\theta)=\frac{l_a}{l_o}=\frac{1}{\frac{4}{3}}=\frac{3}{4}$