Question

If tan θ = $\frac{1}{\sqrt{7}}$, then $\frac{(\cos e{c}^2\mathrm{\theta }-se{c}^2\mathrm{\theta })}{(\cos e{c}^2\mathrm{\theta }+se{c}^2\mathrm{\theta })}=?$ = ?
(a) $\frac{-2}{3}$
(b) $\frac{-3}{4}$
(c) $\frac{2}{3}$
(d) $\frac{3}{4}$

Answer 1

(d) $\frac{3}{4}$
$$\begin{gathered} =\frac{{\mathrm{cosec}}^2\theta -{\sec }^2\theta }{{\mathrm{cosec}}^2\theta +{\sec }^2\theta } \\ =\frac{{\sin }^2\theta \left({\displaystyle \frac{1}{{\sin }^2\theta }}-{\displaystyle \frac{1}{{\cos }^2\theta }}\right)}{{\sin }^2\theta \left({\displaystyle \frac{1}{{\sin }^2\theta }}+{\displaystyle \frac{1}{{\cos }^2\theta }}\right)}[\mathrm{Multiplying}\mathrm{the}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}{\sin }^2\mathrm{\theta }] \\ =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta } \end{gathered}$$
= $\frac{1-{\displaystyle \frac{1}{7}}}{1+{\displaystyle \frac{1}{7}}}=\frac{6}{8}=\frac{3}{4}$