How to simplify :
$$\sqrt{\tan ^2 x + \cot ^2x }$$
the option are :
(i) $ \tan x \cdot \sin x$
(ii) $\sin x \cdot \cos x $
(iii) $ \sec x \cdot \csc x $
(iv) $ \frac{1}{\tan x - \cot x}$
(v) $ \csc^2 x - \sec ^2 x$
My approach :
Since $\tan x = \frac{\sin x}{\cos x}$ and $\cot x = \frac{\cos x}{\sin x}$,
$$ \begin{align} \sqrt{\tan ^2 x + \cot ^2x } &= \sqrt{\frac{\sin ^2 x}{\cos ^2 x} + \frac{\cos ^2 x}{\sin ^2 x}} \\ &= \sqrt{\frac{\sin ^4 x + \cos ^4 x}{\sin ^2 x \cdot \cos ^2 x} }\\ &= \sqrt{\frac{(\sin ^2 x + \cos ^2 x)^2 - 2 \sin x \cdot \cos x}{\sin ^2 x \cdot \cos ^2 x} }\\ &= \sqrt{\frac{1 - 2\sin x \cdot \cos x}{\sin ^2 x \cdot \cos ^2 x} }\\ &= \sqrt{\sec ^2 x \cdot \csc^2 x - 2 \sec x \cdot \csc x}\\ &= \sqrt{\sec x \cdot \csc x ( \sec x \cdot \csc x - 2)}\\ \end{align} $$
from this point, I don't have any idea how should I approach this problem to get another form of this equation available on the option.
Another approach I have in mind is from changing $\cot x = \frac{1}{\tan x}$
$$ \begin{align} \sqrt{\tan ^2 x + \cot ^2x } &= \sqrt{\tan ^2 x + \frac{1}{\tan ^2 x}} \\ &= \sqrt{\frac{\tan ^4 x + 1}{\tan ^2 x} }\\ \end{align} $$
From this point, I don't have any idea.
What am I missing or what approach should you suggest to change the form to the option available on the option?
