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?
Best Answer
You made a mistake in one of your steps. You should have
$$\sin^4x + \cos^4x = (\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$$
This then helps you achieve
$$\sqrt{\sec^2x\csc^2x-2}$$
Other than that, none of the choices are correct (as far as I can tell, the above expression can't be simplified any further).