TA+B+C = 180° then cot
A B C С
-
+ cot + cot
2
2 2
RALD PACKAGE
2 ABC the value of cos(4+28 + 2C)+c8(479)
o
fA+B = 45°, then prove that (1 + tanA) (1 + tanB) = 2.
YA,B,C are angles of a triangle and if none of them is equ
an A + tan B + tan C = tanA.tanB.tanC.
Matrix Match Type:
1) sin150.cos 150
p) 2-1
2.5
cos(a+2b+3c)/2+cos(a-c)/2
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We have, cotA+cotB+cotC=
3
Squaring,
cot
2
A+cot
2
B+cot
2
C+2cotAcotC+2cotBcotC+2cotCcotA=3 ...(1)
Now in △ABC. A+B+C=π⇒A+B=π−C
⇒cot(A+B)=cot(π−C)
⇒
cotA+cotB
cotA.cotB−1
=−cotC
⇒cotAcotB+cotBcotC+cotCcotA=1 ...(2)
From (1) and (2) , we have,
cot
2
A+cot
2
B+cot
2
C+2cotAcotB+2cotBcotC+2cotCcotA
=3[cotAcotB+cotBcotC+cotCcotA]
⇒cot
2
A+cot
2
B+cot
2
C−cotAcotB−cotBcotC−cotCcotA=0
⇒
2
1
[(cotA−cotB)
2
+(cotB−cotC)
2
+(cotC−cotA)
2
]=0
[∵x
2
+y
2
+z
2
−zx−xy−zy=
2
1
{(x−y)
2
+(y−z)
2
+(z−x)
2
}]
⇒(cotA−cotB)
2
+(cotB−cotC)
2
+(cotC−cotA)
2
=0
⇒cotA=cotB=cotC
[Since R.H.S. is zero, each square must be zero]
⇒A=B=C [for triangle]
⇒ Triangle is equilateral.
Hiiiii, hope it is helpful
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