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Given,tan(90°–A) = cot(90°–B).What is the value of sin(A–B) ?
Dear student As tan (90 -A) = cot (90 -B) Cot A = tan BCos A/ sin A - sinB /cos B = 0 Cos (A +B) = 0 Hemce A +B = 90°
Dear student The explanation is incorrect , Cos A/ sin A = sinB /cos B CosA CosB – SinASinB = 0 So , A+ B = 90 A = 90-B A-B = 90-2B Sin(A-B) = Sin (90-2B) Sin(A-B) = Cos2B hope this helps
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