### Instruction: Simplify $\cot 10 + \tan 5.$

Enter your answer as a trigonometric function evaluated at an integer, such as "sin 7".### Response: I see that there are two trigonometric functions with different arguments, so I wonder if there is a way to use some identities or formulas to relate them or simplify them.
One identity that comes to mind is the double angle formula for tangent, which says that $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}.$
Maybe I can use this to rewrite one of the terms as a double angle.
For example, I can write $\cot 10 = \frac{1}{\tan 10} = \frac{1}{\tan (2 \times 5)}.$
Then I can use the double angle formula to get $\cot 10 = \frac{1}{\tan (2 \times 5)} = \frac{1 - \tan^2 5}{2 \tan 5}.$
Now I have both terms in terms of $\tan 5$, so I can combine them by finding a common denominator.
I get $\cot 10 + \tan 5 = \frac{1 - \tan^2 5}{2 \tan 5} + \tan 5 = \frac{1 - \tan^2 5 + 2 \tan^2 5}{2 \tan 5} = \frac{1 + \tan^2 5}{2 \tan 5}.$
This looks simpler than the original expression, but maybe I can simplify it further by using another identity.
I recall that $\tan^2 x + 1 = \sec^2 x$, which is the Pythagorean identity for tangent and secant.
So I can write $\frac{1 + \tan^2 5}{2 \tan 5} = \frac{\sec^2 5}{2 \tan 5}.$
This is almost in the form of a trigonometric function evaluated at an integer, but not quite.
I need to somehow get rid of the fraction and the secant.
I think of another identity that relates secant and cosine, which is $\sec x = \frac{1}{\cos x}.$
So I can write $\frac{\sec^2 5}{2 \tan 5} = \frac{1}{\cos^2 5} \cdot \frac{1}{2 \tan 5}.$
Then I can use the reciprocal identity for tangent, which is $\tan x = \frac{\sin x}{\cos x}$, to get $\frac{1}{\cos^2 5} \cdot \frac{1}{2 \tan 5} = \frac{1}{\cos^2 5} \cdot \frac{\cos 5}{2 \sin 5}.$
Now I can simplify the fraction by canceling out a factor of $\cos 5$ from the numerator and the denominator.
I get $\frac{1}{\cos^2 5} \cdot \frac{\cos 5}{2 \sin 5} = \frac{1}{\cos 5} \cdot \frac{1}{2 \sin 5} = \frac{\sec 5}{2 \sin 5}.$
This is the final simplified expression, and it is in the form of a trigonometric function evaluated at an integer.
So I can enter it as my answer.