Question

1 - cos A+ cos B-cos(A+B)/
1+ cos A-cos B-cos(A+B)

Answer 1

Step-by-step explanation:

take the numerator of L.H.S

1-cosA+cosB-cos(A+B) simplifying 1-cosA = 2sin2 A/2 by half angle identity

cosB-cos(A+B) = 2sin(A/2)sin((A+2B)/2) difference into product identity

2sin2 A/2 + 2sin(A/2)sin((A+2B)/2) by simple trigonometric identities

by further simplifying

2sin(A/2)(sin A/2 + sin((A+2B)/2)

now applying sum to product identity

2sin(A/2)(2sin((A+B)/2)cosB/2-------------1

and then take denominator

1+cosA-cosB-cos(A+B)

similarly by half angle and sum into product identity we get

2cos2 A/2 + 2cos(A/2)cos((A+2B)/2)

by further simplifying

2cos(A/2)(cos A/2 - cos((A+2B)/2)

applying difference to product identity we get

2cos(A/2)(2sin((A+B)/2)sinB/2---------------2

now dividing 1 by 2

we get

tanA/2cotB/2

so proved