In a triangle ABC, if C = 90°, then a2+b2a2−b2sin⁡ (A−B)=

Since C = 90° ⇒ A + B = 90°, we havea2+b2a2−b2sin⁡ (A−B)=sin2⁡ A+sin2⁡ Bsin2⁡ A−sin2⁡ Bsin⁡ (A−B)=sin2⁡ A+sin2⁡ 90∘−Asin ⁡(A+B)sin⁡ (A−B)sin⁡ (A−B)=sin2⁡ A+cos2⁡ Asin⁡ (A+B)=1sin⁡ 90∘=1=sin⁡ (A+B)

In a triangle ABC, if C = 90°, then a2+b2a2−b2sin⁡ (A−B)=

Since C = 90° ⇒ A + B = 90°, we havea2+b2a2−b2sin⁡ (A−B)=sin2⁡ A+sin2⁡ Bsin2⁡ A−sin2⁡ Bsin⁡ (A−B)=sin2⁡ A+sin2⁡ 90∘−Asin ⁡(A+B)sin⁡ (A−B)sin⁡ (A−B)=sin2⁡ A+cos2⁡ Asin⁡ (A+B)=1sin⁡ 90∘=1=sin⁡ (A+B)

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In a triangle ABC, if C = 90°, then $\frac{a^{2} + b^{2}}{a^{2} - b^{2}} \sin (A - B) =$

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sin (A + B)

cos (A + B)

cos (A – B

sin A + sin B

answer is A.

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Detailed Solution

Since C = 90° $\Rightarrow$ A + B = 90°, we have

$\begin{array}{l} \frac{a^{2} + b^{2}}{a^{2} - b^{2}} \sin (A - B) \\ = \frac{\sin^{2} A + \sin^{2} B}{\sin^{2} A - \sin^{2} B} \sin (A - B) \\ = \frac{\sin^{2} A + \sin^{2} (90^{\circ} - A)}{\sin (A + B) \sin (A - B)} \sin (A - B) \\ = \frac{\sin^{2} A + \cos^{2} A}{\sin (A + B)} = \frac{1}{\sin 90^{\circ}} = 1 = \sin (A + B) \end{array}$