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If $$\frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}}{\text{,}}$$ then the value of $$\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$$ = ?
Answer & Solution
Correct Answer:
Option
B
$$\eqalign{
& \frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}} \cr
& \frac{{{x^2}}}{{{y^2}}}{\text{ = }}\frac{{{{\left( {a + 2} \right)}^2}}}{{{{\left( {a - 2} \right)}^2}}} \cr} $$
Applying componendo and dividendo
$$\eqalign{ & \therefore \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} \cr & = \frac{{{{\left( {a + 2} \right)}^2} - {{\left( {a - 2} \right)}^2}}}{{{{\left( {a + 2} \right)}^2} + {{\left( {a - 2} \right)}^2}}} \cr & = \frac{{8a}}{{2{a^2} + 8}} \cr & = \frac{{4a}}{{{a^2} + 4}} \cr} $$
Applying componendo and dividendo
$$\eqalign{ & \therefore \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} \cr & = \frac{{{{\left( {a + 2} \right)}^2} - {{\left( {a - 2} \right)}^2}}}{{{{\left( {a + 2} \right)}^2} + {{\left( {a - 2} \right)}^2}}} \cr & = \frac{{8a}}{{2{a^2} + 8}} \cr & = \frac{{4a}}{{{a^2} + 4}} \cr} $$
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