How to express this derivative using only sine, cosine, and tangent?

Questions
1

f[x_] = Sin - (x - Sqrt[2]/2 x Sin)/Cos
D[f, {x, 1}]

The result obtained after taking the first derivative of the function is as follows:

Cos[x] - Sec[x] (1 - (x Cos[x])/Sqrt[2] - Sin[x]/Sqrt[2]) - 
 Sec[x] (x - (x Sin[x])/Sqrt[2]) Tan[x]

How can the result of the first derivative of this function be expressed solely in terms of sine, cosine, and tangent as follows?

2025-11-06_183459.png
2025-11-06_183459.png441Ɨ124 12.5 KB

2

It is tricky to stop Sec turning up. Here is a way:

Factor[f'[x] // Simplify]
% /. Sec[x] :> 1/Inactive[Cos][x] // Together
3

You might try ComplexExpand[]:

f[x_] := Sin[x] - (x - Sqrt[2]/2 x Sin[x])/Cos[x];
ComplexExpand[D[f[x], {x, 1}]]
4

Thank you for your help. This result is exactly in the format I wanted.

5

Although the obtained result does not contain secant terms, its form is not what I wanted. The desired form is as shown in the image in the original post.

2025-11-09_072658.png

6

Also:

SetSystemOptions[
  "SimplificationOptions" -> "AutosimplifyTrigs" -> False];

Then either of these:

Simplify[D[Sin[x] - (x - Sqrt[2]/2 x Sin[x])/Cos[x], x], 
 ComplexityFunction -> (LeafCount[#] + 
     10 Count[#, _Sec | _Csc | _Cot, Infinity] &)]

D[Sin[x] - (x - Sqrt[2]/2 x Sin[x])/Cos[x], x] // Simplify // 
 ReplaceAll[Sec[t_] :> 1/Cos[t]]

Simplify[] will convert 1/Cos[t] to Sec[t], since Sec[t] is less complex. You have to modify the complexity function to indicate to Simplify[] that you consider Sec[t] more complex than 1/Cos[t].

Or manually substitute the desired forms for the undesired ones. With "AutosimplifyTrigs" turned off, they stay as substituted.