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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Mathematics
Chapter
Heights and Distances

BITSAT 2009 · Mathematics · Heights and Distances

In a triangle (ABC, C=90^ ), then ((a^2-b^2)/(a^2+b^2)) is equal to...

In a triangle (\mathrm{ABC}, \angle \mathrm{C}=90^{\circ}), then (\frac{a^2-b^2}{a^2+b^2}) is equal to :

Options

  1. A(\sin (\mathrm{A}+\mathrm{B}))
  2. B(\sin (A-B))
  3. C(\cos (\mathrm{A}+\mathrm{B}))
  4. D(\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right))

Correct answer

B. (\sin (A-B))

Step-by-step solution

(A+B=180∘−C=90∘a=2Rsin⁡A,b=2Rsin⁡B,c=2Rsin⁡C∴a2−b2a2+b2=sin⁡2 A−sin⁡2 Bsin⁡2 A+sin⁡2 B=sin⁡(A+B)sin⁡(A−B)sin⁡2 A+sin⁡2(90∘−A)[∵A+B=90∘]=sin⁡90∘sin⁡(A−B)sin⁡2 A+cos⁡2 A=sin⁡(A−B)\begin{aligned} & \mathrm{A}+\mathrm{B}=180^{\circ}-\mathrm{C}=90^{\circ} \\ & \mathrm{a}=2 \mathrm{R} \sin \mathrm{A}, b=2 \mathrm{R} \sin \mathrm{B}, c=2 \mathrm{R} \sin \mathrm{C} \\ & \begin{array}{l} \therefore \frac{a^2-b^2}{a^2+b^2}=\frac{\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}}{\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}} \\ \quad=\frac{\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})}{\sin ^2 \mathrm{~A}+\sin ^2\left(90^{\circ}-\mathrm{A}\right)} \\ {\left[\because \mathrm{A}+\mathrm{B}=90^{\circ}\right]} \\ =\frac{\sin 90^{\circ} \sin (\mathrm{A}-\mathrm{B})}{\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}}=\sin (\mathrm{A}-\mathrm{B}) \end{array} \end{aligned})
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