Question

Given $\cos \theta =\frac{q}{\sqrt{p^2+q^2}}$, find $\frac{\csc \theta +\cot \theta }{\csc \theta -\cot \theta }$.

Answer 1

We have,

$\cos \theta =\frac{q}{\sqrt{p^2+q^2}}$

Since,

$\Rightarrow \frac{\csc \theta +\cot \theta }{\csc \theta -\cot \theta }$

$\Rightarrow \frac{\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }}{\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta }}$

$\Rightarrow \frac{1+\cos \theta }{1-\cos \theta }$

On putting the value of $\cos \theta$, we get

$\Rightarrow \frac{1+\frac{q}{\sqrt{p^2+q^2}}}{1-\frac{q}{\sqrt{p^2+q^2}}}$

$\Rightarrow \frac{\sqrt{p^2+q^2}+q}{\sqrt{p^2+q^2}-q}$

$\Rightarrow \frac{\sqrt{p^2+q^2}+q}{\sqrt{p^2+q^2}-q}\times \frac{\sqrt{p^2+q^2}+q}{\sqrt{p^2+q^2}+q}$

$\Rightarrow \frac{{(\sqrt{p^2+q^2}+q)}^2}{{\left(\sqrt{p^2+q^2}\right)}^2-q^2}$

$\Rightarrow \frac{p^2+q^2+q^2+2q\sqrt{p^2+q^2}}{p^2+q^2-q^2}$

$\Rightarrow \frac{p^2+2q^2+2q\sqrt{p^2+q^2}}{p^2}$

Hence, this is the answer.