Question

(1+COSA) (1+cosB)(1+COSC) = (1-ACOSA) (1-cosB) (1-cosC) = ?​

Answer 1

$\huge\boxed{\fcolorbox{red}{yellow}{ANSWER..}}$

Given that - 

   (1-cos a)(1-cos b)(1-cos c)=(1+cos a)(1+ cos b)(1+ cos c)

To prove-

any of the value of the family is sina sinb sinc

case1 

  multiply  both sides by (1-cos a)(1- cos b)(1- cos c)

We get,

[ (1- cos a)(1- cos b)(1- cos c)]² = [sina sin sinc]²

  So we find,

(1- cos a)(1- cos b)(1- cos c) = sina sinb sinc

Case 2

-------  multiply  both sides by  (1+cos a)(1 + cos b)(1 + cos c) 

So we find again, 

(1 + cos a)(1+ cos b)(1+ cos c) = sina sinb sin c

We find that,

any of the member in two cases can be equal  to sina sinb sinc

Answer 2

Hello(｡・//ε//・｡)

Answer:~

Given:-(1+cos a)(1+cos b)(1+cos c)=(1-cos a)(1-cos b)(1-cos c)

Case 1:-

Multiplying both sides by (1-cos a)(1-cos b)(1-cos c) we get,

(1-cos a)(1+cos a)(1-cos b)(1+cos b)(1-cos c)(1+cos c)=((1-cos a)(1-cos b)(1-cos c))^2

=>(1-cos^2a)(1-cos^2b)(1-cos^2c)=((1-cos a)(1-cos b)(1-cos c))^2

=>sin^2a×sin^2b×sin^2c=((1-cos a)(1-cos b)(1-cos c))^2

=>(1-cos a)(1-cos b)(1-cos c)=sin a×sin b×sin c

Case 2:-

Again, on multiplying (1+cos a)(1+cos b)(1+cos c) on both sides,

We get,((1+cos a)(1+cos b)(1+cos c))^2=sin^2a×sin^2b×sin^2c

=>(1+cos a)(1+cos b)(1+cos c)=sin a×sin b×sin c

Therefore, we found that both the sides is equal to sin a×sin b×sin c.

[Hence, proved]

❥Thank You(◍•ᴗ•◍)❤

Itzgirl45⚘