Solve this

Solution:

Let us consider a right $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}$ and $\angle C=\theta$.

Now it is given that $\tan \theta=\frac{A B}{B C}=\frac{1}{\sqrt{7}}$.

So, if $\mathrm{AB}=k$, then $\mathrm{BC}=\sqrt{7} k$, where $k$ is a positive number.

Using Pythagoras theorem, we have:

$A C^{2}=A B^{2}+B C^{2}$

$\Rightarrow \mathrm{AC}^{2}=(k)^{2}+(\sqrt{7} k)^{2}$

$\Rightarrow \mathrm{AC}^{2}=k^{2}+7 k^{2}$

 

$\Rightarrow \mathrm{AC}=2 \sqrt{2} k$

Now, finding out the values of the other trigonometric ratios, we have:

$\sin \theta=\frac{A B}{A C}=\frac{k}{2 \sqrt{2} k}=\frac{1}{2 \sqrt{2}}$

$\cos \theta=\frac{B C}{A C}=\frac{\sqrt{7} k}{2 \sqrt{2} k}=\frac{\sqrt{7}}{2 \sqrt{2}}$

$\therefore \operatorname{cosec} \theta=\frac{1}{\sin \theta}=2 \sqrt{2}$ and $\sec \theta=\frac{1}{\cos \theta}=\frac{2 \sqrt{2}}{\sqrt{7}}$

Substituting the values of $\operatorname{cosec} \theta$ and $\sec \theta$ in the given expression, we get:

$\frac{\cos e c^{2} \theta-\sec ^{2} \theta}{\cos e c^{2} \theta+\sec ^{2} \theta}$

$=\frac{(2 \sqrt{2})^{2}-\left(\frac{2 \sqrt{2}}{\sqrt{7}}\right)^{2}}{(2 \sqrt{2})^{2}+\left(\frac{2 \sqrt{2}}{\sqrt{7}}\right)^{2}}$

$=\frac{8-\left(\frac{8}{7}\right)}{8+\left(\frac{8}{7}\right)}$

$=\frac{\frac{56-8}{7}}{\frac{56+8}{7}}$

$=\frac{48}{64}=\frac{3}{4}=\mathrm{RHS}$

 i.e., LHS = RHS
 
Hence proved.