Prove that sin A cos B + cos A sin B = 1.

If in ∆ABC, ∠C = 90° and tan A = ¾, prove that sin A cos B + cos A sin B = 1.

Answer :

tan A = BC/AC = ¾

AB² = AC² + BC²

= 4² + 3²

= 16 + 9

= 25

= 5²

AB = 5

sin A = BC/AC = 3/5

cos A = AC/AB = 4/5

cos B = BC/AB = 3/5

sin B = AC/AB = 4/5

LHS = sin A cos B + cos A sin B

= (3/5 × 3/5) + (4/5 × 4/5)

= 9/25 + 16/25

= (9 + 16)/ 25

= 25/25

= 1

= RHS

Hence, LHS = RHS.

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