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We have to prove the following equation

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I tried as

Taking square root both side and making denominator same in LHS .

After I got stuck .

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    What's $\vec{a}^2$.? – 2017-01-21
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    If by "taking square root of both sides" you mean just removing the squares, you can't do that here -- it's meaningless. In each set of parentheses you have a vector, and writing a vector squared means taking its dot product with itself: $\vec{a}^2=\vec{a}\cdot\vec{a}$. If you just eliminate the squaring notation, you end up with a couple of vectors, which actually are not the same -- only their "dot-product" squares are. Moreover, there's a useful property that you will need to solve this problem: $\vec{a}^2=|\vec{a}|^2$. – 2017-01-21

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Hint - On solving left hand side

$\left( \frac1{|\vec{a}|^2} + \frac1{|\vec{b}|^2} - 2 \frac{\vec{a}.\vec{b}}{|\vec{a}|^2.|\vec{b}|^2}\right)$

= $\left( \frac{|\vec{a}|^2 + |\vec{b}|^2 - 2 \vec{a}.\vec{b}}{|\vec{a}|^2.|\vec{b}|^2} \right)$

= $\left( \frac{\vec{a} - \vec{b}}{|\vec{a}|.|\vec{b}|} \right)^2$

As $\vec a^2 = |\vec a|^2$