Question

# If $$x=\left(\dfrac{a}{b}\right)^{\dfrac{2ab}{a^{2}-b^{2}}}$$, shew that $$\dfrac{ab}{a^{2}+b^{2}}\left(x^{\dfrac{a}{b}}+x^{\dfrac{b}{a}}\right)=\left(\dfrac{a}{b}\right)^{\dfrac{a^{2}+b^{2}}{a^{2}-b^{2}}}$$

Solution

## Given,$$x=(\dfrac{a}{b})^{\dfrac{2ab}{a^2-b^2}}$$$$(x)^{\dfrac{a}{b}}=(\dfrac{a}{b})^{\dfrac{2a^2}{a^2-b^2}}$$$$(x)^{\dfrac{b}{a}}=(\dfrac{a}{b})^{\dfrac{2b^2}{a^2-b^2}}$$Taking L.H.S.$$\dfrac{ab}{a^2+b^2}(x^{\dfrac{a}{b}}+x^{\dfrac{b}{a}})$$$$\to \dfrac{ab}{a^2+b^2}((\dfrac{a}{b})^{\dfrac{2a^2}{a^2-b^2}}+(\dfrac{a}{b})^{\dfrac{2b^2}{a^2-b^2}})$$$$\to \dfrac{ab}{a^2+b^2}(\dfrac{a^{\dfrac{2a^2}{a^-b^2}}}{b^{\dfrac{2a^2}{a^2-b^2}}}+\dfrac{a^{\dfrac{2b^2}{a^2-b^2}}}{b^{\dfrac{2b^2}{a^2-b^2}}})$$Taking L.C.M$$\to \dfrac{ab}{a^2+b^2}(\dfrac{a^{\dfrac{2a^2}{a^2-b^2}}.b^{\dfrac{2b^2}{a^2-b^2}}+a^{\dfrac{2b^2}{a^2-b^2}}.b^{\dfrac{2a^2}{a^2-b^2}}}{b^{\dfrac{2a^2+2b^2}{a^2-b^2}}})$$Multiplying ab inside$$\to \dfrac{1}{a^2+b^2}(\dfrac{a^{\dfrac{3a^2-b^2}{a^2-b^2}}.b^{\dfrac{a^2+b^2}{a^2-b^2}}+a^{\dfrac{a^2+b^2}{a^2-b^2}}.b^{\dfrac{3a^2-b^2}{a^2-b^2}}}{b^{\dfrac{2a^2+2b^2}{a^2-b^2}}})$$Taking common$$\to \dfrac{a^{\dfrac{a^2+b^2}{a^2-b^2}}.b^{\dfrac{a^2+b^2}{a^2-b^2}}}{a^2+b^2}(\dfrac{a^{\dfrac{2a^2-2b^2}{a^2-b^2}}+b^{\dfrac{2a^2-2b^2}{a^2-b^2}}}{b^{\dfrac{2a^2+2b^2}{a^2-b^2}}})$$$$\to \dfrac{a^{\dfrac{a^2+b^2}{a^2-b^2}}}{a^2+b^2}(\dfrac{a^2+b^2}{b^{\dfrac{a^2+b^2}{a^2-b^2}}})$$$$\to (\dfrac{a}{b})^{\dfrac{a^2+b^2}{a^2-b^2}}$$=R.H.S Mathematics

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