Question

if 0°<A<90°, prove that sinA+cosA>1.​

Answer 1

Answer:

Solved and proved below.

Step-by-step explanation:

Let A be 30° , 45° , 60°.

Case 1: 30°

sinA + cosA

= $\frac{1}{2} + \frac{\sqrt{3} }{2}$       > 1

Case 2: 45°

sinA + cosA

=    $\frac{1}{\sqrt{2} } + \frac{1}{\sqrt{2} }$  

= $\sqrt{2}$    >1

Case 3: 60°

sinA + cosA

= $\frac{\sqrt{3} }{2} + \frac{1}{2}$       > 1

Hence proved.