Question

If tan θ = $\frac{1}{\sqrt{7}}$ then prove that $\left(\frac{{\mathrm{cosec}}^2\mathrm{\theta }+{\sec }^2\mathrm{\theta }}{{\mathrm{cosec}}^2\mathrm{\theta }-{\sec }^2\mathrm{\theta }}\right)=\frac{4}{3}.$

Answer 1

Let us consider a right $△$ABC, right angled at B and $\angle C=\theta$.
Now it is given that tan $\theta$ = $\frac{AB}{BC}$ = $\frac{1}{\sqrt{7}}$.

So, if AB = k, then BC = $\sqrt{7}$k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AC² = (k)² + ($\sqrt{7}$k)²
⇒ AC² = k² + 7k²
⇒ AC = 2$\sqrt{2}$k
Now, finding out the values of the other trigonometric ratios, we have:
sin $\theta$ = $\frac{AB}{AC}=\frac{k}{2\sqrt{2}k}=\frac{1}{2\sqrt{2}}$
cos $\theta$ = $\frac{BC}{AC}=\frac{\sqrt{7}k}{2\sqrt{2}k}=\frac{\sqrt{7}}{2\sqrt{2}}$
∴ cosec $\theta$ = $\frac{1}{\sin \theta }=2\sqrt{2}$ and sec $\theta$ = $\frac{1}{\cos \theta }=\frac{2\sqrt{2}}{\sqrt{7}}$
Substituting the values of cosec $\theta$ and sec $\theta$ in the given expression, we get:
$$\begin{gathered} \frac{\cos e{c}^2\theta -se{c}^2\theta }{\cos e{c}^2\theta +se{c}^2\theta } \\ =\frac{(2\sqrt{2}{)}^2-{\left({\displaystyle \frac{2\sqrt{2}}{\sqrt{7}}}\right)}^2}{(2\sqrt{2}{)}^2+{\left({\displaystyle \frac{2\sqrt{2}}{\sqrt{7}}}\right)}^2} \\ =\frac{8-\left({\displaystyle \frac{8}{7}}\right)}{8+\left({\displaystyle \frac{8}{7}}\right)} \\ =\frac{{\displaystyle \frac{56-8}{7}}}{{\displaystyle \frac{56+8}{7}}} \\ =\frac{48}{64}=\frac{3}{4}=\mathrm{RHS} \end{gathered}$$
i.e., LHS = RHS

Hence proved.