Question

If $\frac{3a+7b}{3a-7b}=\frac{4}{3}$ then find the value of $\frac{3a^2-7b^2}{3a^2+7b^2}$

Answer 1

$\frac{3a+7b}{3a-7b}=\frac{4}{3}$

Applying componendo and dividendo

$\frac{3a+7b+3a-7b}{3a+7b-3a+7b}=\frac{4+3}{4-3}$

Solving we get $\frac{a}{b}=\frac{49}{3}$

So, $\frac{a^2}{b^2}=\frac{2401}{9}$

$\frac{3a^2}{7b^2}=\frac{343}{3}$

Applying componendo and dividendo , $\frac{3a^2+7b^2}{3a^2-7b^2}=\frac{343+3}{343-3}=\frac{346}{340}=\frac{173}{170}$

So $\frac{3a^2-7b^2}{3a^2+7b^2}=\frac{170}{173}$