Question

What is the crt ans for this qn???​

Answer 1

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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