Question

Eliminate $\theta$ from the following.
$a=\sin \theta -\frac{1}{\sin \theta },b=\cos \theta -\frac{1}{\cos \theta }$

Answer 1

We have,

$a=\sin \theta -\frac{1}{\sin \theta }$

On squaring both sides, we get

$a^2={\sin }^2\theta +\frac{1}{{\sin }^2\theta }-2$ $...........\left(1\right)$

Similarly,

$b^2={\cos }^2\theta +\frac{1}{{\cos }^2\theta }-2$ $...........\left(2\right)$

On adding both equations, we get

$a^2+b^2={\sin }^2\theta +{\cos }^2\theta +\frac{1}{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta }-2-2$

$a^2+b^2=1+\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }-4$

$a^2+b^2=\frac{1}{{\sin }^2\theta {\cos }^2\theta }-3$ $............\left(3\right)$

Since,

$a=\frac{{\sin }^2\theta -1}{\sin \theta }$

$a=\frac{-{\cos }^2\theta }{\sin \theta }$

$\frac{1}{a}=-\frac{\sin \theta }{{\cos }^2\theta }$

$\frac{1}{a^2}=\frac{{\sin }^2\theta }{{\cos }^4\theta }$ $...............\left(4\right)$

Similarly,

$\frac{1}{b^2}=-\frac{{\cos }^2\theta }{{\sin }^4\theta }$ $...............\left(5\right)$

On multiplying both equations, we get

$\frac{1}{a^2{b}^2}=\frac{1}{{\sin }^2\theta {\cos }^2\theta }$ $.............\left(6\right)$

From equations $\left(3\right)$ and $\left(6\right)$

$a^2+b^2=\frac{1}{a^2{b}^2}-3$

Hence, this is the answer.