Given sin(−θ)=1/5 and tanθ=√6/12 . what is the value of cosθ ? a). √6/60 b). −2√6/5 c). −√6/60 d). 2√6/5

B). −2√6/5

Step-by-step explanation:

tan theta = sin theta/ cos theta

Multiply each side by cos theta

tan theta * cos theta = sin theta

Divide each side by tan theta

cos theta = sin theta/ tan theta

We know that the sin (- theta) = - sin theta  since sin is and odd function

sin theta  = - (  sin (-theta))

Putting this into the above equation,

cos theta =  - (  sin (-theta)) / tan theta

cos theta = - 1/5 / (sqrt(6)/12)

Remember when dividing fractions, we use copy dot flip

cos theta =   -1/5  * 12/ sqrt(6)

cos theta = -12/ (5 sqrt(6))

We cannot leave a sqrt in the denominator, so multiply the top and bottom by sqrt(6)/sqrt(6)

cos theta = -12/ (5 sqrt(6))  * sqrt(6)/sqrt(6)

cos theta = -12 sqrt(6) / 5*6

Simplify the fraction.

cos theta = -2 sqrt(60/5

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length