TS EAMCET · Maths · Properties of Triangles
In a \(\triangle A B C, C=90^{\circ}\). Then, \(\frac{a^2-b^2}{a^2+b^2}\) is equal to
- A \(\sin (A+B)\)
- B \(\sin (A-B)\)
- C \(\cos (A+B)\)
- D \(\cos (A-B)\)
Answer & Solution
Correct Answer
(B) \(\sin (A-B)\)
Step-by-step Solution
Detailed explanation
\(\angle C=90^{\circ}, \frac{a^2-b^2}{a^2+b^2}\) ...(i) \(\because \quad \cos c=\frac{a^2+b^2-c^2}{2 a b}=\cos 90^{\circ}\) \(\Rightarrow \quad \frac{a^2+b^2-c^2}{2 a b}=0\) \(a^2+b^2-c^2=0\) \(\Rightarrow \quad c^2=a^2+b^2\) From Eq. (i), we get…
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