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2015-04-19T15:25:40+05:30

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By triangle sum poperty we have
A+B+C=180
Since B is a right angle
A+90+C=180
C=90-A

Now consider the given expression
\sin(A)+\cos(A)=\sin(A)+\sin(90-A)\\=\sin(A)+\sin(C)\\=\frac{a}{b}+\frac{c}{b}\\=\frac{a+c}{b}\\ >\frac{b}{b}\\=1
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ty :)