`f(x)=(cos^2x)/(1+cosx+cos^2x)` and `g(x)=ktanx+(1-k)sinx-x`, where `k in R, g'(x)=`

Options

A. `((1-cos x ) (f (x)-k))/(cos x)`
B. `((1- cos x )(k-f (x)))/(cos x)`
C. `((1- cos x )(k -f (x)))/(f(x))`
D. `((1- cos x )(k -f (x)))/((f(x ))^(2))`

Answer

To solve the problem, we need to find the derivative \( g'(x) \) of the function \( g(x) = k \tan x + (1-k) \sin x - x \).

### Step-by-Step Solution:

1. **Identify the Function**:
\[ g(x) = k \tan x + (1-k) \sin x - x \]

2. **Differentiate \( g(x) \)**:
We will differentiate \( g(x) \) with respect to \( x \).
- The derivative of \( \tan x \) is \( \sec^2 x \).
- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of \( x \) is \( 1 \).

Therefore, we have:
\[ g'(x) = k \sec^2 x + (1-k) \cos x - 1 \]

3. **Combine Terms**:
We can rewrite \( g'(x) \):
\[ g'(x) = k \sec^2 x + (1-k) \cos x - 1 \]

4. **Find a Common Denominator**:
To combine the terms, we can express \( \sec^2 x \) in terms of \( \cos x \):
\[ \sec^2 x = \frac{1}{\cos^2 x} \]
Thus,
\[ g'(x) = \frac{k}{\cos^2 x} + (1-k) \cos x - 1 \]

5. **Combine into a Single Fraction**:
Multiply the second term by \( \cos^2 x \) to have a common denominator:
\[ g'(x) = \frac{k + (1-k) \cos^3 x - \cos^2 x}{\cos^2 x} \]

6. **Factor the Numerator**:
The numerator can be factored using the identity for the difference of cubes:
\[ 1 - \cos^3 x = (1 - \cos x)(1 + \cos x + \cos^2 x) \]
Thus, we can write:
\[ g'(x) = \frac{k(1 - \cos^3 x) + (1-k)(1 - \cos^2 x)}{\cos^2 x} \]

7. **Final Expression**:
After simplifying, we arrive at:
\[ g'(x) = \frac{(k - (1-k)) (1 - \cos x)(1 + \cos x + \cos^2 x)}{\cos^2 x} \]

### Final Result:
Thus, the derivative \( g'(x) \) is:
\[ g'(x) = \frac{(k - 1)(1 - \cos x)(1 + \cos x + \cos^2 x)}{\cos^2 x} \]