Question

If A is an acute angle in right triangle ABC, right angled at B, then prove that sin A + cos A > 1

Answer 1

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
$\sin(A)+\cos(A)=\sin(A)+\sin(90-A)\\=\sin(A)+\sin(C)\\=\frac{a}{b}+\frac{c}{b}\\=\frac{a+c}{b}\\ >\frac{b}{b}\\=1$