The first question sinα=4/5, cos(α+β)=5/ 13.

Then cos α = √ [1-(4/5) 2] = 3/5.

sin(α+β)=√[ 1-(5/ 13)^2]= 12/ 13

Then sin β = sin [(α+β)-α] = sin (α+β) cos α-cos (α+β) sin α = (12/13) * (3/5)-(5/13).

The second question sinβ= number of roots1010, so cos β = number of roots10/0.

sin2β=2sinβcosβ=3/5

Sinβ= root number 10/ 10 < 1/2.

So β < 30, so 2β is also an acute angle.

So cos2β=4/5.

tan2β=3/4

tan(α+2β)=(tanα+tan 2β)/( 1-tanα* tan 2β)

=( 1/7+3/4)/[ 1-( 1/7)(3/4)]

=(25/28)/( 1-3/28)

= 1

Question 4 (1) Prove from right to left

Right = (1-tan α tan β) tan (α+β) = (1-tan α tan β) * (tan α+tan β)/(1-tan α tan β) = tan α+tan β = left.

(2)tan 60 = tan(20+40)=(tan 20+tan 40)/( 1-tan 20 tan 40)

tan60=√3

Namely: (tan 20+tan 40)/(1-tan 20 tan 40) = √ 3.

tan 20+tan 40 =√3( 1-tan 20 tan 40)

tan20+tan40=√3-√3tan20tan40

tan20+tan40+√3tan20tan40=√3

(3) The original formula = 1+tanα*tanβ-(tanα+tanβ).

= 2-[(tanα+tanβ)+ 1-tanα* tanβ]

= 2-( 1-tanα* tanβ)( 1+tan(α+β))

=2-0

=2

(4) tan 60 = (tan 20+tan 40)/(1-tan 20 * tan 40) = root number 3.

Tan20+tan40 = root number 3 *( 1-tan20 * tan40)

Substitute into the above formula

(tan 20+tan 40+tan 120) divided by (tan20 tan40)

= (root number 3- root number 3 * tan 20 * tan 40- root number 3)/(tan20 * tan40)

=-root number 3