Question

In triangle ABC angle C=90 degrees then prove that sin(A-B) = a^2-b^2/a^2+b^2

Answer 1

Given : In triangle ABC angle C=90 degrees

Step-by-step explanation:

Sin(A-B) = SinACosB - CosASinB

Triangle is right angle triangle at C

SinA  =  a/c

CosA = b/c

SinB =  b/c

CosB = a/c

=> Sin(A-B) = (a/c)(a/c) - (b/c)(b/c)

=> Sin(A-B) = a²/c² - b²/c²

=> Sin(A-B) = (a² - b²)/c²

using Pythagoras theorem

c²  = a² +  b²

=> Sin(A-B) = (a² - b²)/(a² +  b²)

QED

Hence Proved

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