Question

Show that , x-1+y-1\x-1+x-1-y-1\x-1=x2+y2\xy

Answer 1

Given : (x⁻¹  + y⁻¹)/x⁻¹    + (x⁻¹  - y⁻¹)/y⁻¹  = (x² + y²)/xy

To find : Prove the equality

Solution:

Correction in Question :   other wise LHS = 2

(x⁻¹  + y⁻¹)/x⁻¹    + (x⁻¹  - y⁻¹)/y⁻¹  = (x² + y²)/xy

LHS

= (x⁻¹  + y⁻¹)/x⁻¹    + (x⁻¹  - y⁻¹)/y⁻¹

= (1/x + 1/y)/(1/x)  + (1/x - 1/y)/(1/y)

= x(y + x) /xy   +  y(y - x)/xy

= (1/xy) ( xy + x²   + y²  - yx)

= (1/xy) ( x² + y²)

=  (x² + y²)/xy

= RHS

QED

Hence Proved

(x⁻¹  + y⁻¹)/x⁻¹    + (x⁻¹  - y⁻¹)/y⁻¹  = (x² + y²)/xy

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Answer 2

Answer:

Concept:

For the equation to be valid, the numerical value of the expression on the left side of the equation must be equal to the numerical value of the variable on the right side of the equation under all conditions.

Step-by-step explanation:

Given:

$\left(x^{-1}+y^{-1}\right) / x^{-1}+\left(x^{-1}-y^{-1}\right) / y^{-1}=\left(x^{2}+y^{2}\right) / x y$

Find:

Prove that LHS is equal to RHS

Solution:

Taking LHS

$\begin{aligned}&=\left(x^{-1}+y^{-1}\right) / x^{-1}+\left(x^{-1}-y^{-1}\right) / y^{-1} \\&=(1 / x+1 / y) /(1 / x)+(1 / x-1 / y) /(1 / y) \\&=x(y+x) / x y+y(y-x) / x y \\&=(1 / x y)\left(x y+x^{2}+y^{2}-y x\right) \\&=(1 / x y)\left(x^{2}+y^{2}\right) \\&=\left(x^{2}+y^{2}\right) / x y \\&=R H S\end{aligned}$

Hence proved.

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