Question

Let 0 be an angle in quadrant 3 such that cos0=-5/13. Find the exact values of csc0 and tan0

Answer 1

Answer:

see explanation

Step-by-step explanation:

Given

cosθ = - $\frac{5}{13}$

This is a 5- 12- 13 right triangle

with adjacent = 5, opposite = 12 and hypotenuse = 13

In quadrant 3

sinθ < 0 and tanθ > 0 , then

sinθ = $\frac{opposite}{hypotenuse}$ = -  $\frac{12}{13}$ and

cscθ = $\frac{1}{sin0}$ = $\frac{1}{-\frac{12}{13} }$ = - $\frac{13}{12}$

then

tanθ = $\frac{sin0}{cos0}$ = $\frac{-\frac{12}{13} }{-\frac{5}{13} }$ = - $\frac{12}{13}$ × - $\frac{13}{5}$ = $\frac{12}{5}$