In any ΔABC, prove that\(\frac{cosA}a\) + \(\frac{cosB}b\) + \(\f

1.	In any ΔABC, prove that\(\frac{cosA}a\) + \(\frac{cosB}b\) + \(\frac{cosc}c\) = \(\frac{(a^2+b^2+c^2)}{2abc}\)
Answer» Need to prove: \(\frac{cosA}a\) + \(\frac{cosB}b\) + \(\frac{cosc}c\) = \(\frac{(a^2+b^2+c^2)}{2abc}\) Left hand side = \(\frac{cosA}a\) + \(\frac{cosB}b\) + \(\frac{cosc}c\) = \(\frac{b^2+c^2-a^2}{2abc}\) + \(\frac{c^2+a^2-b^2}{2abc}\) + \(\frac{a^2+b^2-c^2}{2abc}\) = \(\frac{a^2+b^2+c^2}{2abc}\) = Right hand side. [proved]

In any ΔABC, prove that\(\frac{cosA}a\) + \(\frac{cosB}b\) + \(\frac{cosc}c\) = \(\frac{(a^2+b^2+c^2)}{2abc}\)

Answer»

Need to prove: \(\frac{cosA}a\) + \(\frac{cosB}b\) + \(\frac{cosc}c\) = \(\frac{(a^2+b^2+c^2)}{2abc}\)

Left hand side

= \(\frac{cosA}a\) + \(\frac{cosB}b\) + \(\frac{cosc}c\)

= \(\frac{b^2+c^2-a^2}{2abc}\) + \(\frac{c^2+a^2-b^2}{2abc}\) + \(\frac{a^2+b^2-c^2}{2abc}\)

= \(\frac{a^2+b^2+c^2}{2abc}\)

= Right hand side. [proved]