xy = a+2a−2,xy = a+2a−2,\frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}}{\text{,}} then the value of x2−y2x2+y2" role="presentation">x2−y2x2+y2x2−y2x2+y2\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} = ?" /> xy = a+2a−2,xy = a+2a−2,\frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}}{\text{,}} then the value of x2−y2x2+y2" role="presentation">x2−y2x2+y2x2−y2x2+y2\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} = ?" /> xy = a+2a−2,xy = a+2a−2,\frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}}{\text{,}} then the value of x2−y2x2+y2" role="presentation">x2−y2x2+y2x2−y2x2+y2\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} = ?" />

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If $\frac{x}{y} = \frac{a + 2}{a - 2},$ $\frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}}{\text{,}}$ then the value of $\frac{x^{2} - y^{2}}{x^{2} + y^{2}}$ $\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$ = ?

1 $\frac{2 a}{a^{2} + 2}$ $\frac{{2a}}{{{a^2} + 2}}$
2 $\frac{4 a}{a^{2} + 4}$ $\frac{{4a}}{{{a^2} + 4}}$
3 $\frac{2 a}{a^{2} + 4}$ $\frac{{2a}}{{{a^2} + 4}}$
4 $\frac{4 a}{a^{2} + 2}$ $\frac{{4a}}{{{a^2} + 2}}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{x}{y} = \frac{a + 2}{a - 2} \\ \frac{x^{2}}{y^{2}} = \frac{{(a + 2)}^{2}}{{(a - 2)}^{2}} \end{aligned}

$\eqalign{ & \frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}} \cr & \frac{{{x^2}}}{{{y^2}}}{\text{ = }}\frac{{{{\left( {a + 2} \right)}^2}}}{{{{\left( {a - 2} \right)}^2}}} \cr}$
Applying componendo and dividendo

\begin{aligned} ∴ \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \\ = \frac{{(a + 2)}^{2} - {(a - 2)}^{2}}{{(a + 2)}^{2} + {(a - 2)}^{2}} \\ = \frac{8 a}{2 a^{2} + 8} \\ = \frac{4 a}{a^{2} + 4} \end{aligned}

$\eqalign{ & \therefore \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} \cr & = \frac{{{{\left( {a + 2} \right)}^2} - {{\left( {a - 2} \right)}^2}}}{{{{\left( {a + 2} \right)}^2} + {{\left( {a - 2} \right)}^2}}} \cr & = \frac{{8a}}{{2{a^2} + 8}} \cr & = \frac{{4a}}{{{a^2} + 4}} \cr}$

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then the value of

\frac{x^{2} - y^{2}}{x^{2} + y^{2}}

$\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$ = ?", "text": "If

\frac{x}{y} = \frac{a + 2}{a - 2},

$\frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}}{\text{,}}$ then the value of

\frac{x^{2} - y^{2}}{x^{2} + y^{2}}

$\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$ = ?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{x}{y} = \frac{a + 2}{a - 2} \\ \frac{x^{2}}{y^{2}} = \frac{{(a + 2)}^{2}}{{(a - 2)}^{2}} \end{aligned}

$\eqalign{ & \frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}} \cr & \frac{{{x^2}}}{{{y^2}}}{\text{ = }}\frac{{{{\left( {a + 2} \right)}^2}}}{{{{\left( {a - 2} \right)}^2}}} \cr}$
Applying componendo and dividendo

\begin{aligned} ∴ \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \\ = \frac{{(a + 2)}^{2} - {(a - 2)}^{2}}{{(a + 2)}^{2} + {(a - 2)}^{2}} \\ = \frac{8 a}{2 a^{2} + 8} \\ = \frac{4 a}{a^{2} + 4} \end{aligned}

$\eqalign{ & \therefore \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} \cr & = \frac{{{{\left( {a + 2} \right)}^2} - {{\left( {a - 2} \right)}^2}}}{{{{\left( {a + 2} \right)}^2} + {{\left( {a - 2} \right)}^2}}} \cr & = \frac{{8a}}{{2{a^2} + 8}} \cr & = \frac{{4a}}{{{a^2} + 4}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{x}{y} = \frac{a + 2}{a - 2} \\ \frac{x^{2}}{y^{2}} = \frac{{(a + 2)}^{2}}{{(a - 2)}^{2}} \end{aligned}

$\eqalign{ & \frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}} \cr & \frac{{{x^2}}}{{{y^2}}}{\text{ = }}\frac{{{{\left( {a + 2} \right)}^2}}}{{{{\left( {a - 2} \right)}^2}}} \cr}$
Applying componendo and dividendo

\begin{aligned} ∴ \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \\ = \frac{{(a + 2)}^{2} - {(a - 2)}^{2}}{{(a + 2)}^{2} + {(a - 2)}^{2}} \\ = \frac{8 a}{2 a^{2} + 8} \\ = \frac{4 a}{a^{2} + 4} \end{aligned}

$\eqalign{ & \therefore \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} \cr & = \frac{{{{\left( {a + 2} \right)}^2} - {{\left( {a - 2} \right)}^2}}}{{{{\left( {a + 2} \right)}^2} + {{\left( {a - 2} \right)}^2}}} \cr & = \frac{{8a}}{{2{a^2} + 8}} \cr & = \frac{{4a}}{{{a^2} + 4}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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