if A+B+C=pi/2 prove that?

Question

(cosA+sinB+sinC)
divided by (sinA+cosB+sinC)
=(1-tanA/2)/((1+tan A/2)

Answer 1

for simplicty say A = 2x, B= 2y, C= 2z

A+B+C = pi/2
x+y+z = pi/4
y+z = pi/4-x
take tan of both sides
tan(y+z) = tan(pi/4-x)

using tan(a+b) = (tan a + tan b)/(1-tan a tan b)
we get
(tan y + tan z)/(1- tan y tan z) = (1-tan x)/(1+tan x) ... 1
now rhs is rhs of given expression as x = A/2
we need to show the LHS of your expression
numeratior of your expression
(Cos 2x + sin 2y+ sin 2z) = sin(2y+2z) as 2x+2y+2z = pi/2
= sin 2y cos 2z + sin 2z cos 2y + sin 2y + sin 2z
= sin 2y(cos 2z+1) + sin 2z(cos 2y+1)
= sin 2y cos ^2 z + sin 2z cos ^2 y
= 2 siny cos y cos ^2 z + 2 sin z cos z cos ^2 y
= 2 cos ^z cos ^y (tan y + tan z) putting sin y = cos y tan y and taking common factor out ...2

denomenator =
sin 2x + cos 2y + sin 2z
= cos(2y+2z) + cos2y + sin 2z
= cos2y cos2z - sin 2y sin 2z + cos 2y + sin 2z
= cos2y(cos 2z+ 1) + sin 2z(1- sin 2y)
= cos 2y cos^2 z - 2 sin z cos z ( 1- 2 sin y cos y)
= cos^2 z(cos 2y - 2 tanz(1-2 siny cos y))


from the above you should be able to reduce it and by division you should be able to get the resulty

Answer 2

First thing... think about what it means for a+b+c = pi/2.

Answer 3

Solve left hand side, and then right hand side. Both sides of equation are xactly euqal

Answer 4

go reverse!