Question

If tan $\mathrm{\theta }=\sqrt{3},\mathrm{then}\frac{{\sec }^2\mathrm{\theta }-{\mathrm{cosec}}^2\mathrm{\theta }}{{\sec }^2\mathrm{\theta }+{\mathrm{cosec}}^2\mathrm{\theta }}\mathit{=}\mathit{?}$
(a) −1
(b) 1
(c) $-\frac{1}{2}$
(d) $\frac{1}{2}$

Answer 1

(d) $\frac{1}{2}$
Here,
$\tan \theta =\sqrt{3}$
$\Rightarrow \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\sqrt{3}}{1}$
By Pythagoras' theorem, we get AB = 1.

Now, $\mathrm{sec\theta }=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{2}{1}$ and $\mathrm{cosec\theta }=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{2}{\sqrt{3}}$
$$\begin{gathered} \frac{se{c}^2\theta -\cos e{c}^2\theta }{se{c}^2\theta +\cos e{c}^2\theta } \\ =\frac{{\left(2\right)}^2-{\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}^2}{{\left(2\right)}^2+{\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}^2} \\ =\frac{4-{\displaystyle \frac{4}{3}}}{4+{\displaystyle \frac{4}{3}}} \\ =\frac{12-4}{12+4} \\ =\frac{8}{16} \\ =\frac{1}{2} \end{gathered}$$