請幫我做幾條maths.. about trigon架

1)tan17。+ tan 13。 / tangent 17。tan13。 - 1
。即係degree 咁既意思

因为
tan (A+B) = tan(A) + tan(B)
-----------------------------
1 – tan(A)tan(B)
所以
tan17°+ tan 13°/ tangent 17° tan13° - 1
= - (tan17°+ tan 13° / 1 - tangent 17°tan13°)
= - tan(17°+ 13°)
= - tan(30°) = -1/sqrt(3)

2) cos (2B + pie/6) cos (2B-pie/6) +sin(2B+pie /6)sin(2B-pie/6)
因为
cos(A-B) = cosAcosB + sinAsinB
所以
cos (2B + pie/6) cos (2B-pie/6) +sin(2B+pie /6)sin(2B-pie/6)
= cos((2B + pie/6) – (2B – pie/6))
= cos(pie/3)
= 1/2

2)Given that tanA= 3/4, tan B = 12/5, find the value of tan (A-B)

tan (A-B) = tan(A) - tan(B)
-------------------------------------
1 + tan(A)tan(B)
= (3/4 – 12/5) / (1 + (3/4)(12/5))
= - 33 / 56

tan (A+B) = (tan(A) + tan(B))/(1 – tan(A)tan(B))
so
(tan17°+ tan 13°)/ tangent 17° tan13° - 1
= - (tan17°+ tan 13° / 1 - tangent 17°tan13°)
= - tan(17°+ 13°)
= - tan(30°) = -1/sqrt(3)

2) cos(A-B) = cosAcosB + sinAsinB
so
cos (2B + pie/6) cos (2B-pie/6) +sin(2B+pie /6)sin(2B-pie/6)
= cos((2B + pie/6) – (2B – pie/6))
= cos(pie/3)
= 1/2

3)tan (A-B) = (tan(A) - tan(B))/(1 + tan(A)tan(B))
= (3/4 – 12/5) / (1 + (3/4)(12/5))
= - 33 / 56

1) tan17。+ tan 13。 / tangent 17。tan13。 - 1

0.536598872/-0.92941651