Consider this equation. sin(θ)= 3sqrt10/10

If θ is an angle in quadrant II, what is the value of tan(θ) ?

tan x = -3 (Answer A)

Step-by-step explanation:

We want to find the tangent of this angle "theta," and recall the trig identity

(sin x)^2 + (cos x)^2 = 1.

               3√10

If sin x =

                   10

                             90

then (sin x)^2 = = 9/10

                             100

and (cos x)^2 = 1 - 9/10 = 1/10

                      sin x       3√10/10

Then tan x = = = -3 (Answer A)

                       cos x      1√10/10

The tangent function is negative in Quadrant II.  In Quadrant I tan x = +3

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92