InterviewSolution
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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. |
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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 |
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