Question

Sinx = 1/2, cosy = sqrt2/2, and angle x and angle y are both in the first quadrant.

tan(x+y)=

A. -3.73
B. 1.53
C. 3.00
D. 3.73​

Answer 1

D. 3.73

Explanation:

Answer 2

Option D. 3.73​

Explanation:

we know that

`tan(x+y)=(tan(x)+tan(y))/(1-tan(x)tan(y))`

and

`sin^(2)(\alpha)+cos^(2)(\alpha)=1`

step 1

Find cos(X)

we have

`sin(x)=(1)/(2)`

we know that

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

substitute

`((1)/(2))^(2)+cos^(2)(x)=1`

`cos^(2)(x)=1-(1)/(4)`

`cos^(2)(x)=(3)/(4)`

`cos(x)=(√(3))/(2)`

step 2

Find tan(x)

`tan(x)=sin(x)/cos(x)`

substitute

`tan(x)=1/√(3)`

step 3

Find sin(y)

we have

`cos(y)=(√(2))/(2)`

we know that

`sin^(2)(y)+cos^(2)(y)=1`

substitute

`sin^(2)(y)+((√(2))/(2))^(2)=1`

`sin^(2)(y)=1-(2)/(4)`

`sin^(2)(y)=(2)/(4)`

`sin(y)=(√(2))/(2)`

step 4

Find tan(y)

`tan(y)=sin(y)/cos(y)`

substitute

`tan(y)=1`

step 5

Find tan(x+y)

`tan(x+y)=(tan(x)+tan(y))/(1-tan(x)tan(y))`

substitute

`tan(x+y)=[1/√(3)+1}]/[{1-1/√(3)}]=3.73`