# How to simplify $\sqrt{\tan^2 x + \cot^2x }$?

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?

• Use \tan, \sin and \cos etc. to produce $\tan$, $\sin$ and $\cos$ instead of $tan$, $sin$ and $cos$ which look bad.
– Gary
Oct 18, 2021 at 3:19
• @Gary thanks for the finishing touches Oct 18, 2021 at 3:22
• The value of the question at $x=\dfrac{\pi}{4}$ isn't satisfied anywhere in the options.
– UNAN
Oct 18, 2021 at 3:31
• @PCMSE Yeah I plot on GeoGebra I thought maybe I'm missing something I doubt the options.
– user960916
Oct 18, 2021 at 3:33

$$\sin^4x + \cos^4x = (\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$$
$$\sqrt{\sec^2x\csc^2x-2}$$