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Answer 1 · 2017-02-09T08:33:28

mathworld.wolfram.com

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

So by Pythgorean theorem

#AB^2=AC^2+BC^2=a^2+a^2=2a^2#

#=>AB=sqrt2a#

Again #/_ABC=/_BAC=45^@#

So

#cot45-sin45=cotA-sinA#

#="adjacent"/"opposite"-"opposite"/"hypotnuse"#

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

#=a/a-a/(sqrt2a)#

#=1-1/sqrt2#

#=1-sqrt2/2#

#=(2-sqrt2)/2#