This is the basic and conceptual question from trigonometric ratios in which we have been asked to prove that 2sin 45° cos 45° = sin 90°
RS Aggarwal, Class 10, chapter 11, question no 11(iv).
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L.H.S. = 2 sin 45° cos 45°
= 2 × (1/√2) × (1/√2)
= (2 × 1/2) = 1
R.H.S. = sin 90° = 1
L.H.S. = R.H.S.