Question #f4fb1

1 Answer
Feb 9, 2017

mathworld.wolfram.commathworld.wolfram.com

Let us consider a right angled isosceles triangle ABC inwhich /_ACB="rtangle" and AC=BC=a(say)ACB=rtangleandAC=BC=a(say)

So by Pythgorean theorem

AB^2=AC^2+BC^2=a^2+a^2=2a^2AB2=AC2+BC2=a2+a2=2a2

=>AB=sqrt2aAB=2a

Again /_ABC=/_BAC=45^@ABC=BAC=45

So

cot45-sin45=cotA-sinAcot45sin45=cotAsinA

="adjacent"/"opposite"-"opposite"/"hypotnuse"=adjacentoppositeoppositehypotnuse

=(AC)/(BC)-(BC)/(AB)=ACBCBCAB

=a/a-a/(sqrt2a)=aaa2a

=1-1/sqrt2=112

=1-sqrt2/2=122

=(2-sqrt2)/2=222