Question #d8ac3

2 Answers
Oct 17, 2017

See the answer below...

Explanation:

#(1+cosA)(1+cosB)(1+cosC)=(1-cosA)(1-cosB)(1-cosC)=k# [Assume it]

Now we can write that
#k^2={(1+cosA)(1+cosB)(1+cosC)}{(1-cosA)(1-cosB)(1-cosC)}#
#=>k^2=(1-cos^2A)(1-cos^2B)(1-cos^2C)#
#=>k^2=sin^2A.sin^2B.sin^2C#
#=>k=+-sinAsinBsinC#

As we assumed before #(1+cosA)(1+cosB)(1+cosC)=k=+-sinAsinBsinC#
PROVED

Hope it helps...
Thank you

Please refer to a Proof in the Explanation.

Explanation:

We will use the following Identities :

#(1) : 1+cos2x=2cos^2x.#

#(2) : 1-cos2x=2sin^2x.#

#(3) : sin2x=2sinxcosx.#

Given that,

#(1+cosA)(1+cosB)(1+cosC)=(1-cosA)(1-cosB)(1-cosC),#

#:. (2cos^2(A/2))(2cos^2(B/2))(2cos^2(C/2))=(2sin^2(A/2))(2sin^2(B/2))(2sin^2(C/2),#

#:. cos^2(A/2)/sin^2(A/2)*cos^2(B/2)/sin^2(B/2)*cos^2(C/2)/sin^2(C/2)=1.#

#:. cot^2(A/2)cot^2(B/2)cot^2(C/2)=1.#

#:. cot(A/2)cot(B/2)cot(C/2)=+-1.............(star).#

Now, #(1+cosA)/sinA=(2cos^2(A/2))/(2sin(A/2)cos(A/2))=cot(A/2).#

Similarly,

#(1+cosB)/sinB=cot(B/2), and, (1+cosC)/sinC=cot(C/2).#

#"Therefore, "(1+cosA)/sinA*(1+cosB)/sinB*(1+cosC)/sinC,#

#=cot(A/2)cot(B/2)cot(C/2),#

#=pm1.............[because, (star)].#

#rArr (1+cosA)(1+cosB)(1+cosC)=+-sinAsinBsinC.#

Hence, the Proof.