Question

Given sin0=6/11 and sec 0<0 find cos0 and tan0 (picture provided)

Answer 1

Answer:

$cos(0\°)=-\frac{\sqrt{85}}{11}$

$tan(0\°) = -\frac{6}{\sqrt{85}}$

Step-by-step explanation:

We know by definition that:

$cos ^ 2(x) = 1-sin ^2(x)$

We also know that:

$tan(x) = \frac{sin(x)}{cos(x)}$

$sec(x) = \frac{1}{cos(x)}$

Then we can use these identities to solve the problem

if $sin(0\°) = \frac{6}{11}$ then $cos^2(0\°) = 1-(\frac{6}{11})^2$

$cos(0\°) = \±\sqrt{\frac{85}{121}}\\\\cos(0\°)=\±\frac{\sqrt{85}}{11}$

As $sec(x) <0$ then $cos(x) <0$. Therefore we take the negative root:

$cos(0\°)=-\frac{\sqrt{85}}{11}$

Now that we know $sin(0\°)$ and $cos(0\°)$ we can find $tan(0\°)$

$tan(0\°) = \frac{sin(0\°)}{cos(0\°)}\\\\tan(0\°) = \frac{\frac{6}{11}}{\frac{-\sqrt{85}}{11} }\\\\tan(0\°) = -\frac{6}{\sqrt{85}}$