Question

Solve these questions with proper explanation. :)

Answer 1

As A, B, C are the angles of triangle
So, A+B+C =180°
A+B= 180°- C
1) Now, sin (A+B) = sin (180°-C)
[By, replacing (A+B) by (180°-C)]

Then, sin (180°-C) = cos (180°-C-90°) = cos (90°-C)
(Using sin A = cos 90°-A)

Cos 90°-C = Sin C
(As Sin A = Cos 90°-A)
Sin C = R.H.S
L.H.S = R.H.S
Hence proved


2) {Cos (A+B)}/2 = {Cos (180°-C)}/2
{Replacing (A+B) by (180°-C)}
We get, cos 90°-C/2
Then, sin C/2
(Using sin A = Cos 90°-A)
So, sin C/2 = R.H.S
L.H.S= R.H.S
Hence proved


Hope this helps you