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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

$cot45-sin45=cotA-sinA$

$="adjacent"/"opposite"-"opposite"/"hypotnuse"$

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

$=a/a-a/(sqrt2a)$

$=1-1/sqrt2$

$=1-sqrt2/2$

$=(2-sqrt2)/2$