Question

How do you simplify \[\cot x + \tan x\]?

Answer 1

Hint:
In the given question, we have been given an expression. This expression is the sum of two different trigonometric functions. We have to simplify the value of the sum of the two trigonometric functions. To do that, we are going to convert the values into their rudimentary forms. Then, we are going to add the two trigonometric functions as the sum of the rudimentary forms. Then we are going to apply the basic trigonometric formulae to them, make them equal to their result, and then simplify the answer.

Complete step by step answer:
The given expression is \[p = \cot x + \tan x\].
Now, \[\cot x = \dfrac{{\cos x}}{{\sin x}}\] and \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
Hence, \[p = \dfrac{{\cos x}}{{\sin x}} + \dfrac{{\sin x}}{{\cos x}}\]
Taking the LCM and solving,
\[p = \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{\sin x\cos x}}\]
Now, we know that \[{\sin ^2}x + {\cos ^2}x = 1\]
Hence, \[p = \dfrac{1}{{\sin x\cos x}}\]
We know the formula of \[\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\]
Hence, \[\sin 2x = \sin x\cos x + \sin x\cos x = 2\sin x\cos x\]
or, \[\sin x\cos x = \dfrac{{\sin 2x}}{2}\]
Thus, \[p = \dfrac{2}{{\sin 2x}}\]
or, \[p = 2\csc 2x\]

Hence, \[\cot x + \tan x = 2\csc 2x\]

Additional Information:
Here, we had the answer of \[2\csc \left( {2x} \right)\], but in the previous steps, using them, we could have arrived at a different answer too,
\[p = \dfrac{1}{{\sin x\cos x}}\]
And since, \[\dfrac{1}{{\sin x}} = \csc x\] and \[\dfrac{1}{{\cos x}} = \sec x\], we could say,
\[p = \csc x\sec x\]

Note:
In the given question, we were given the sum of two trigonometric functions. We had to simplify them. We did that by first converting them into their basic forms, then we added them as the converted forms, applied the known formulae, substituted their results, and that is how we got our answer.