MHT CET · Maths · Properties of Triangles
In any \(\triangle \mathrm{ABC}\), with usual notations, \(\mathrm{c}(\mathrm{a} \cos \mathrm{B}-\mathrm{b} \cos \mathrm{A})=\)
- A \(a^2-b^2\)
- B \(\frac{1}{\mathrm{a}^2}-\frac{1}{\mathrm{~b}^2}\)
- C \(a^2+b^2\)
- D \(\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}\)
Answer & Solution
Correct Answer
(A) \(a^2-b^2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & c(a \cos B-b \cos A) \\ & =a c\left(\frac{c^2+a^2-b^2}{2 a c}\right)-b c\left(\frac{b^2+c^2-a^2}{2 b c}\right) \\ & =\frac{c^2+a^2-b^2}{2}-\frac{b^2+c^2-a^2}{2}=\frac{2 a^2-2 b^2}{2}=a^2-b^2\end{aligned}\)
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