1.

In a triangle ABC, if \(\frac{sin(A-B)}{sin(A+B)}=\frac{a^2-b^2}{a^2+b^2},\) Prove that the triangle is either isosceles or right angled.

Answer»

Given,

\(\frac{sin(A-B)}{sin(A+B)}=\frac{a^2-b^2}{a^2+b^2}\)

Applying componendo and dividendo

\(\frac{sin(A-B)+sin(A+B)}{sin(A-B)-sin(A+B)} \)\(=\frac{a^2-b^2+a^2+b^2}{a^2-b^2-a^2-b^2}\)

⇒ \(-\frac{2\,sin\,A\,cos\,B}{2\,cos\,A\,sin\,B}=\frac{2a^2}{-2b^2}\)

⇒ \(\frac{sin\,A\,cos\,B}{cos\,A\,sin\,B}=\frac{a^2}{b^2}\)

⇒ b2 sin A cos B = a2 cos A sin B

Putting sin A = ka, sin B = kb

⇒ b2ka cos B = a2kb cos A

⇒ b cos B = a cos A

Applying cosine formula,

\(b(\frac{a^2+c^2-b^2}{2ac})\) \(a(\frac{b^2+c^2-b^2}{2bc})\)

⇒ b2 (a2 + c2 – b2 ) = a2 (b2 + c2 – a2)

⇒ a2b2 + b2c2 – b4 = a2b2 + a2c2 – a4

⇒ b2c2 – a2c2 – b4 + a4 = 0

⇒ c2 (b2 – a2) – (b2 – a2 ) (b2 + a2 ) = 0

⇒ (b2 – a2) (c2 – b2 – a2 ) = 0

⇒ Either b2 – a2 = 0 

or, c2 – b2 – a2 = 0

⇒ b = a 

or, c2 = b2 + a2

∴ Triangle is isosceles if b = a

∴ Triangle is right angled if c2 = b2 + a2



Discussion

No Comment Found