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Let 0 be an angle in quadrant 3 such that cos0=-5/13.

Find the exact values of csc0 and tan0

Let 0 Be An Angle In Quadrant 3 Such That Cos0513 Find The Exact Values Of Csc0 And Tan0 class=

Sagot :

Answer:

see explanation

Step-by-step explanation:

Given

cosθ = - [tex]\frac{5}{13}[/tex]

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 = [tex]\frac{opposite}{hypotenuse}[/tex] = -  [tex]\frac{12}{13}[/tex] and

cscθ = [tex]\frac{1}{sin0}[/tex] = [tex]\frac{1}{-\frac{12}{13} }[/tex] = - [tex]\frac{13}{12}[/tex]

then

tanθ = [tex]\frac{sin0}{cos0}[/tex] = [tex]\frac{-\frac{12}{13} }{-\frac{5}{13} }[/tex] = - [tex]\frac{12}{13}[/tex] × - [tex]\frac{13}{5}[/tex] = [tex]\frac{12}{5}[/tex]

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