If a+b+c=180 then prove that cot^2 A/2+cos^2 B/2+cos^2C/2=2+2sinA/2sinB/2sinC/2
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Answer:
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.
Step-by-step explanation:
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