Question

Given that:
(1 + cosA)(1 + cosB)(1 + cosC) = (1 - cosA)(1 - cosB)(1 - cosC)

Show that one of the values of each member of this equality is SinA.SinB.SinC

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Answer 1

$\huge\underline\pink{Given:-}$

• (1+cosA) (1+cosB) (1+cosC) =(1-cosA) (1-cosB) (1-cosC)

$\huge\underline\pink{To Show:-}$

• One of the values of each member of this equality is SinA. SinB. SinC.

$\huge\underline\pink{Solution:-}$

→ (1+cosA) (1+cosB) (1+cosC) =(1-cosA) (1-cosB) (1-cosC).

Multiplying both sides by (1-cosA) (1-cosB) (1-cosC) we get,

→ Sin²A.Sin²B.Sin²C=[(1-cosA) (1-cosB) (1-cosC) ]²

Now, Taking Sqaureroot on both the sides we get,

SinA.SinB.SinC=(1-cosA) (1-cosB) (1-cos)

$\huge\red{Hence Shown}$

$\huge\boxed{\dag\sf\purple{@ItzSiddhi}\dag}$

Answer 2

Hola mate

Here is your answer -

Plz refer to the attachment....