Question

Given that:(1 + cos alpha)(1 + cos beta)(1 + cos gamma ) = (1 - cos alpha)(1 - cos beta)(1 - cos gamma).Show that one of the values of each member of this quality is (sin alpha × sin beta × sin gamma).

Answer 1

Multiply both sides by (1-cosA)(1-cosB)(1-cosC). It makes both the sides (1-cosA)²(1-cosB)²(1-cosC)²=sin²Asin²Bsin²C.

So, the (1-cosA)(1-cosB)(1-cosC)-sinAsinBsinC=0 or (1-cosA)(1-cosB)(1-cosC)+sinAsinBsinC=0. Hence proved.

The value of each member of this equality is considering as sin X and sin y and Sin Z has been evaluated by squaring on both sides.

It has formula for evaluating the tangent of a half angle is given by sin and cos.

Answer 2

Answer:

See the attachment . If you can't understand any step let me know in the comment box.