[Shirsendu Roy]

plz upload actual question, because this question needs modification

considering the question to be find the value of the given expression,

BC/AH+CA/BH+AB/CH

here BC/AH=(AD/AH)(DC/AD)(BC/DC)

=(sinB)(sinA/cosA)(1/sinB)

=tanA

same way CA/BH=tanB

AB/CH=tanC

or, tanA+tanB+tanC=tanAtanBtanC

[where 0=tan(A+B+C)=(tanA+tanB+tanC-tanAtanBtanC)/(1-tanAtanB-tanAtanC-tanBtanC)]

=(BC/AH)(CA/BH)(AB/CH)

 

 

[Shirsendu Roy]

limx tends to infinity F(x) tends to finite .

[Shirsendu Roy]

It is bounded and converges to 0

[Shirsendu Roy]

AH=2RcosA, BH=2RcosB, CH=2RcosC

(AH+BH+CH)=2R(cosA+cosB+cosC)

=2R(1+r/R)

=2(R+r)

<=2R+R since r<=R/2 for any triangle

=3R

2R<=(AH+BH+CH)<=3R

[Shirsendu Roy]

since angle HBD=angle CAD

and BH=AC

the two right triangle triangle BHD and triangle ACD are congruent

then HD=CD or angle HCD=45

CH is perpendicular AB then

angle B +angle HCD=90

then angleB=45

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