Question

If cos(theta) = 5/0 and is in the 1st quadrant, find the following:

Answer 1

step 1

Find out sine

Remember the identity

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

substitute the given value of cosine

$\begin{gathered} (\frac{5}{9})^2+s\imaginaryI n^2\theta=1 \\ \\ s\imaginaryI n^2\theta=1-\frac{25}{81} \\ \\ s\mathrm{i}n^2\theta=\frac{56}{81} \\ \\ sin\theta=\frac{\sqrt{56}}{9} \\ \\ sin\theta=\frac{2\sqrt{14}}{9} \end{gathered}$

step 2

Find out cosecant

$\begin{gathered} csc\theta=\frac{1}{sin\theta} \\ \\ csc\theta=\frac{9}{2\sqrt{14}}*\frac{\sqrt{14}}{\sqrt{14}}=\frac{9\sqrt{14}}{28} \\ \\ csc\theta=\frac{9\sqrt{14}}{28} \end{gathered}$

step 3

Find out secant

$\begin{gathered} sec\theta=\frac{1}{cos\theta} \\ \\ sec\theta=\frac{9}{5} \end{gathered}$

step 4

Find out tangent

$\begin{gathered} tan\theta=\frac{sin\theta}{cos\theta} \\ \\ tan\theta=\frac{\frac{2\sqrt{14}}{9}}{\frac{5}{9}}=\frac{2\sqrt{14}}{5} \end{gathered}$

step 5

Find out cotangent

$\begin{gathered} cot\theta=\frac{1}{tan\theta} \\ \\ cot\theta=\frac{5}{2\sqrt{14}}*\frac{\sqrt{14}}{\sqrt{14}}=\frac{5\sqrt{14}}{28} \\ \\ cot\theta=\frac{5\sqrt{14}}{28} \end{gathered}$