What is the crt ans for this qn???
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A Is a positive acute angle
means, 0° < A < 90° .
now, sin A+ cosA
= √2 [ sin A× 1/√2 + cos A× 1/√2 ]
= √2[sinAcos45° + cosA.sin45° ]
we know, sinA.cosB + cosA.sinB = sin(A + B)
so, [sincos45° + cos.sin45° ] = sin( A+ 45°)
then, sin + cos = √2sin(A+45°)
here, 0° < < 90°
so, sin(A+45°) > sin45° = 1/√2
so, √2sin( A+ 45°) > √2 × 1/√2 = 1
hence, √2sin( A+ 45°) > 1
hence proved
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