If $$\frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}}{\text{,}}$$ then the value of $$\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$$ = ?
A. $$\frac{{2a}}{{{a^2} + 2}}$$
B. $$\frac{{4a}}{{{a^2} + 4}}$$
C. $$\frac{{2a}}{{{a^2} + 4}}$$
D. $$\frac{{4a}}{{{a^2} + 2}}$$
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & \frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}} \cr & \frac{{{x^2}}}{{{y^2}}}{\text{ = }}\frac{{{{\left( {a + 2} \right)}^2}}}{{{{\left( {a - 2} \right)}^2}}} \cr} $$Applying componendo and dividendo
$$\eqalign{ & \therefore \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} \cr & = \frac{{{{\left( {a + 2} \right)}^2} - {{\left( {a - 2} \right)}^2}}}{{{{\left( {a + 2} \right)}^2} + {{\left( {a - 2} \right)}^2}}} \cr & = \frac{{8a}}{{2{a^2} + 8}} \cr & = \frac{{4a}}{{{a^2} + 4}} \cr} $$
Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
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B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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