Question

If 3x=sec$\theta$ and $\frac{3}{x}=tan\theta ,$ then find the value of 9$(x^2-\frac{1}{x^2})$

Answer 1

${\sec }^2\theta -{\tan }^2\theta =1$

$(3x{)}^2-{\left(\frac{3}{x}\right)}^2=1$

$9x^2-\frac{9}{x^2}=1$

$9(x^2-\frac{1}{x^2})=1$