Question

If $2x=sec\theta$ and $\frac{2}{x}=tan\theta$ then $2(x^2-\frac{1}{x^2})$ = ?

(a) $\frac{1}{2}$ (b) $\frac{1}{4}$ (c) $\frac{1}{8}$ (d) $\frac{1}{16}$

Answer 1

Given: $2x=sec\theta \Rightarrow x=\frac{sec\theta }{2}$

$\frac{2}{x}=tan\theta \Rightarrow \frac{1}{x}=\frac{tan\theta }{2}$

Now

$2[x^2-\frac{1}{x^2}]=2[x^2-(\frac{1}{x}{)}^2]=2[(\frac{sec\theta }{2}{)}^2-(\frac{tan\theta }{2}{)}^2]=2[\frac{se{c}^2\theta }{4}-\frac{ta{n}^2\theta }{4}]=\frac{2}{4}[se{c}^2\theta -ta{n}^2\theta ]=\frac{1}{2}\times 1=\frac{1}{2}$

(a) $\frac{1}{2}$