Question

Prove that: $\dfrac{{{a^{ - 1}}}}{{{a^{ - 1}} + {b^{ - 1}}}} + \dfrac{{{a^{ - 1}}}}{{{a^{ - 1}} - {b^{ - 1}}}} = \dfrac{{ - \left( {2{b^2}} \right)}}{{{a^2} - {b^2}}}$

Answer 1

Hint – In this particular question use the concept that solve the unsimplified side of the given equation rather than the simplified form as the unsimplified equations do not have any closed form solution later on in the solution use the concept that ${x^{ - 1}} = \dfrac{1}{x}$, $\dfrac{{\dfrac{x}{y}}}{{\dfrac{p}{q}}}$ = $\dfrac{{xq}}{{yp}}$ and $\left( {p - q} \right)\left( {p + q} \right) = {p^2} - {q^2}$ so use these properties to get the solution of the question.

Complete step-by-step answer:
Given equation:
$\dfrac{{{a^{ - 1}}}}{{{a^{ - 1}} + {b^{ - 1}}}} + \dfrac{{{a^{ - 1}}}}{{{a^{ - 1}} - {b^{ - 1}}}} = \dfrac{{ - \left( {2{b^2}} \right)}}{{{a^2} - {b^2}}}$
Proof –
Consider the LHS of the given equation we have,
$ \Rightarrow \dfrac{{{a^{ - 1}}}}{{{a^{ - 1}} + {b^{ - 1}}}} + \dfrac{{{a^{ - 1}}}}{{{a^{ - 1}} - {b^{ - 1}}}}$
Now solve the above equation using the property the ${x^{ - 1}} = \dfrac{1}{x}$ so apply this in the above equation we have,
$ \Rightarrow \dfrac{{\dfrac{1}{a}}}{{\dfrac{1}{a} + \dfrac{1}{b}}} + \dfrac{{\dfrac{1}{a}}}{{\dfrac{1}{a} - \dfrac{1}{b}}}$
Now first solve the denominator of both the terms we have,
$ \Rightarrow \dfrac{{\dfrac{1}{a}}}{{\dfrac{{b + a}}{{ab}}}} + \dfrac{{\dfrac{1}{a}}}{{\dfrac{{b - a}}{{ab}}}}$
Now use the property that $\dfrac{{\dfrac{x}{y}}}{{\dfrac{p}{q}}}$ is written as $\dfrac{{xq}}{{yp}}$ so apply this in the above equation we have,
$ \Rightarrow \dfrac{{ab}}{{a\left( {b + a} \right)}} + \dfrac{{ab}}{{a\left( {b - a} \right)}}$
Now simplify the above equation by canceling out the common terms in numerator and the denominator we have,
$ \Rightarrow \dfrac{b}{{\left( {b + a} \right)}} + \dfrac{b}{{\left( {b - a} \right)}}$
Now take the LCM of the above equation we have,
$ \Rightarrow \dfrac{{b\left( {b - a} \right) + b\left( {b + a} \right)}}{{\left( {b + a} \right)\left( {b - a} \right)}}$
Now first simplify the numerator of the above equation we have,
$ \Rightarrow \dfrac{{{b^2} - ab + {b^2} + ab}}{{\left( {b + a} \right)\left( {b - a} \right)}}$
Now cancel out the positive and negative same terms from the numerator of the above equation we have,
$ \Rightarrow \dfrac{{{b^2} + {b^2}}}{{\left( {b + a} \right)\left( {b - a} \right)}}$
Now solve the denominator of the above equation using the property which is given as, $\left( {p - q} \right)\left( {p + q} \right) = {p^2} - {q^2}$ so we have,
$ \Rightarrow \dfrac{{{b^2} + {b^2}}}{{{b^2} - {a^2}}} = \dfrac{{2{b^2}}}{{{b^2} - {a^2}}}$
Now take negative one common from the denominator of the above equation we have,
\[ \Rightarrow \dfrac{{2{b^2}}}{{ - \left( { - {b^2} + {a^2}} \right)}}\]
Now take this negative sign to the numerator we have,
\[ \Rightarrow \dfrac{{ - \left( {2{b^2}} \right)}}{{{a^2} - {b^2}}}\]
= RHS
Hence proved.

Note – In such types of questions it is advised to simplify the LHS or the RHS according to their complexity of given function. Sometimes proving LHS = RHS needs simplification on both sides of the equation. Remember to convert dissimilar functions to get to the final result, and check whether R.H.S is equal to L.H.S or not, we will get the required result.