(a+b+c)^2/(a+b+c)=(cotA/2+cotB/2+cotC/2)/(cotA+cotB+cotC)
Ishaan , 9 Years ago
Grade 11
Trigonometry> (a+b+c)^2/(a+b+c)=(cotA/2+cotB/2+cotC/2)/...
1 AnswersYatin Modi
Last Activity: 6 Years ago
Proof:- Numerator= √s(s-a)/√(s-b)(s-c) + √s(s-b)/√(s-a)(s-c) + √s(s-c)/√(s-a)(s-b)
After taking LCM we get,
s(s-a)+s(s-b)+s(s-c)/√s(s-a)(s-b)(s-c)
= s[3s-(a+b+c)/∆
We know that a+b+c = 2s, so
s²/∆= numerator= [(a+b+c)/2]²
And denominator= cotA+ cotB+ cotC
= CosA/sinA + cosB/sinB + cosC/sinC
By cosine formula and sinA=2∆/bc , sinB=2∆/ac, sinC = 2∆/ab, we get
1/4∆ ×[a²+b²+c²] = denominator.
Now, Numerator/ denominator= (a+b+c)²/(4∆×a²+b²+c²)/4∆
= (a+b+c)²/a²+b²+c²
Therefore, hence proved.
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