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TS EAMCET · Maths · Properties of Triangles

If \(\triangle A B C\) is a non isosceles triangle and \(\angle C=90^{\circ}\), then \(\frac{a^2+b^2}{a^2-b^2} \sin (A-B)=\)

  1. A \(1\)
  2. B \(2\)
  3. C \(0\)
  4. D \(-1\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1\)

Step-by-step Solution

Detailed explanation

Given that, in \(\triangle A B C\), \(\begin{aligned} \angle C & =90^{\circ} \\ \angle A+\angle B+\angle C & =180^{\circ} \\ \therefore \quad \angle A+\angle B & =90^{\circ}\end{aligned}\) Then, \(\frac{a^2+b^2}{a^2-b^2} \sin (A-B)\) Using sine rule,we get…