Question

Let 0 be an angle such that sec0= -13/12 and cot0<0. Find the exact values of tan0 and sin0.

Answer 1

Answer:

tan 0 = -2.4

sin 0 = 0.42

Step-by-step explanation:

sec 0 = -13/12 => cos 0 = -12/13

cot 0 < 0 => tan 0 < 0

so, the angle is on second quadrant

=> tan 0 = -12/5 = -2.4

=> sin 0 = 5/12 = 0.42

Answer 2

Answer:

Solution given;

Sec θ=-$\frac{13}{12}$

cotθ< 0,

It lies in second quadrant.

where sin and cosec is positive.

Now

$\frac{1}{cosθ}=-\frac{13}{12}$

cosθ=$\frac{12}{13}$

$\frac{b}{h}$=$\frac{12}{13}$

b=12

h=13

By using Pythagoras law

p=$\sqrt{13²-12²}=5$

Now

exact values of tan θ=$\frac{p}{b}$=$\frac{5}{12}$

since it lies in II quadrant

tan θ=-$\frac{5}{12}$

and

sinθ=$\frac{p}{h}$=$\frac{5}{13}$

since it lies in II quadrant

sin θ=$\frac{5}{13}$