Question

In any triangle ABC, prove that
(c) sin (A-B) =(a²-b²)/c² sinC​

Answer 1

Answer:

note: the "2" is square here.

Step-by-step explanation:

a2−b2/a2+b2=sin(A−B)/sin(A+B)

⇒a2+b2/a2−b2=sin(A+B)/sin(A−B)

Using componendo and dividendo,

⇒2a2/2b2=sin(A+B)+sin(A−B)/sin(A+B)−sin(A−B)

⇒a2/b2=2sinAcosB/2sinBcosA

⇒a2/b2=sinAcosB/sinBcosA

Now, from sine law,

a/sinA=b/sinB=c/sinC=k

⇒a=k sinA,b=k sinB,c=k sinC

So, our equation becomes,

a2/b2=akcosB/bkcosA

⇒cosB/cosA=a/b

⇒b cos B=a cos A

⇒b(a2+c2−b2/ac)=a(b2+c2−a2/bc)

⇒b2a2+b2c2−b4=a2b2+a2c2−a4

⇒c2(b2−a2)=b4−a4

⇒c2(b2−a2)−(b2+a2)(b2−a2)=0

⇒(b2−a2)(c2−(b2+a2))=0

⇒(b2−a2)=0or(c2−(b2+a2))=0

⇒b2=a2orc2=(b2+a2)

⇒b=aorc2=(b2+a2)

It means, either the ΔABC is a right angle triangle or a isoceles triangle.

hope it helps :)