Question

Need to check if my answer for number 20 is correct… i got 192/756 or 7/27

Answer 1

Find out

`tan(\alpha+\beta)`

Remember that

`tan(\alpha+\beta)=\frac{tan\alpha\text{+}tan\beta}{1-tan\alpha *tan\beta}`

step 1

Find out the cosine of angle alpha

the angle alpha lies on the II quadrant

the value of the cosine of angle alpha is negative

Remember that

`\begin{gathered} sin^2\alpha+cos^2\alpha=1 \\ sin\alpha=(3)/(5) \end{gathered}`

substitute

`\begin{gathered} ((3)/(5))^2+cos^2\alpha=1 \\ cos^2\alpha=1-(9)/(25) \\ \\ cos^2\alpha=(16)/(25) \\ cos\alpha=-(4)/(5) \end{gathered}`

Find out the tangent of angle alpha

`tan\alpha=((3)/(5))/(-(4)/(5))=-(3)/(4)`

step 2

Find out the value of the sine of the angle beta

The angle beta lies on the III quadrant

so

The value of the sine is negative

Remember that

`\begin{gathered} sin^2\beta+cos^2\beta=1 \\ cos\beta=-(12)/(13) \end{gathered}`

substitute

`\begin{gathered} s\imaginaryI n^2\beta+(-(12)/(13))^2=1 \\ \\ s\imaginaryI n^2\beta=1-(144)/(169) \\ \\ s\imaginaryI n^2\beta=(25)/(169) \\ \\ s\imaginaryI n\beta=-(5)/(13) \end{gathered}`

Find out the value of the tangent of the angle beta

`tan\beta=(-(5)/(13))/(-(12)/(13))=(5)/(12)`

step 3

Find out

`tan(\alpha+\beta)=\frac{tan\alpha+\text{t}an\beta}{1-tan\alpha tan\beta}`

substitute given values

`tan(\alpha+\beta)=((-(3)/(4))+((5)/(12)))/(1-(-(3)/(4))((5)/(12)))`
`tan(\alpha+\beta)=((-3\/4)+((5)/(12)))/(1+15\/48))=((-9+5)/(12))/((48+15)/(48))=((-4)/(12))/((33)/(48))=-(4)/(12)/(33)/(48)=-(48*4)/(33*12)=-(48)/(99)`

simplify

The answer is -16/33