Question

If x is an acute angle and $\tan x=\frac{1}{\sqrt{7}}$, then the value of $\frac{{\mathrm{cosec}}^{2}x-{\sec }^{2}x}{{\mathrm{cosec}}^{2}x+{\sec }^{2}x}$ is
(a) 3/4
(b) 1/2
(c) 2
(d) 5/4

Answer 1

(a) 3/4

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \tan x=\frac{1}{\sqrt{7}} \\ \therefore {\tan }^{2}x=\frac{1}{7} \\ \mathrm{Now},\mathrm{dividing}\mathrm{the}\mathrm{numerator}\mathrm{and}\mathrm{the}\mathrm{denominator}\mathrm{of}\frac{{\mathrm{cosec}}^{2}x-{\sec }^{2}x}{{\mathrm{cosec}}^{2}x+{\sec }^{2}x}\mathrm{by}{\mathrm{cosec}}^{2}x: \end{gathered}$$
$$\begin{gathered} \frac{1-{\tan }^{2}x}{1+{\tan }^{2}x} \\ =\frac{1-{\displaystyle \frac{1}{7}}}{1+\frac{1}{7}} \\ =\frac{6}{8}=\frac{3}{4} \end{gathered}$$