AP EAMCET · Maths · Properties of Triangles
In \(\triangle \mathrm{ABC}\) if \(\left\lfloor C=90^{\circ}\right.\), then \(\frac{\left(\sin ^2 A+\sin ^2 B\right)}{\left(\sin ^2 A-\sin ^2 B\right)} \sin (A-B)=\)
- A \(1\)
- B \(\frac{1}{2}\)
- C \(\frac{\sqrt{3}}{2}\)
- D \(0\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
Given \(C=90^{\circ}\), then \(A+B=90^{\circ}\). \(\sin^2 A + \sin^2 B = \sin^2 A + \sin^2(90^{\circ}-A) = \sin^2 A + \cos^2 A = 1\) \(\sin^2 A - \sin^2 B = \sin(A-B)\sin(A+B) = \sin(A-B)\sin(90^{\circ}) = \sin(A-B)\)…
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