Question

cos^2A+cos^2B+cos^2C=1-2cosA.cosB.cosC​

Answer 1

Step-by-step explanation:

We write cos

2

A=1−sin

2

A

and as in ΔABC A+B+C=180

⟹cosC=cos(180−A−B)=−cos(A+B)

L.H.S.=1−sin

2

A+cos

2

B+cos

2

C

=1+(cos

2

B−sin

2

A)+cos

2

C

=1+cos(B+A)cos(B−A)+cos

2

C ..... (cos

2

C−sin

2

D=cos(C+D)cos(C−D))

=1−cosCcos(B−A)+cos

2

C

=1−cosC[cos(B−A)−cosC]

=1−cosC[cos(B−A)+cos(B+A)] .

=1−cosC(2cosAcosB) ....... By compound angles formula

=1−2cosAcosBcosC.

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