(e) \( \frac{1-\tan x}{1+\tan x}=\frac{1-2 \sin ^{2} x}{1+2 \sin x \cos x} \)

It seems that the equation is too complex for direct computation. Let's break it down step by step to simplify and solve it manually. We start with the equation: \[ \frac{1 - \tan x}{1 + \tan x} = \frac{1 - 2 \sin^2 x}{1 + 2 \sin x \cos x} \] ### Step 1: Rewrite \(\tan x\) Recall that \(\tan x = \frac{\sin x}{\cos x}\). We can substitute this into the left side of the equation: \[ \frac{1 - \frac{\sin x}{\cos x}}{1 + \frac{\sin x}{\cos x}} = \frac{1 - 2 \sin^2 x}{1 + 2 \sin x \cos x} \] ### Step 2: Simplify the Left Side Multiply the numerator and denominator by \(\cos x\): \[ \frac{\cos x - \sin x}{\cos x + \sin x} \] ### Step 3: Rewrite the Right Side The right side can be simplified as follows: \[ 1 - 2 \sin^2 x = \cos 2x \quad \text{(using the double angle identity)} \] \[ 1 + 2 \sin x \cos x = 1 + \sin 2x \quad \text{(using the double angle identity)} \] Thus, the right side becomes: \[ \frac{\cos 2x}{1 + \sin 2x} \] ### Step 4: Set the Two Sides Equal Now we have: \[ \frac{\cos x - \sin x}{\cos x + \sin x} = \frac{\cos 2x}{1 + \sin 2x} \] ### Step 5: Cross Multiply Cross multiplying gives us: \[ (\cos x - \sin x)(1 + \sin 2x) = \cos 2x(\cos x + \sin x) \] ### Step 6: Expand Both Sides Expanding both sides: Left Side: \[ \cos x + \cos x \sin 2x - \sin x - \sin x \sin 2x \] Right Side: \[ \cos 2x \cos x + \cos 2x \sin x \] ### Step 7: Rearranging the Equation Now we can rearrange the equation to isolate terms involving \(x\): \[ \cos x + \cos x \sin 2x - \sin x - \sin x \sin 2x - \cos 2x \cos x - \cos 2x \sin x = 0 \] ### Step 8: Solve for \(x\) This equation can be quite complex to solve directly. We can look for specific values of \(x\) that satisfy the equation, such as \(x = 0\), \(x = \frac{\pi}{4}\), etc. Let's check if \(x = 0\) is a solution: \[ \frac{1 - \tan(0)}{1 + \tan(0)} = \frac{1 - 2 \sin^2(0)}{1 + 2 \sin(0) \cos(0)} \] \[ \frac{1 - 0}{1 + 0} = \frac{1 - 0}{1 + 0} \Rightarrow 1 = 1 \quad \text{(True)} \] Now let's check \(x = \frac{\pi}{4}\): \[ \frac{1 - \tan(\frac{\pi}{4})}{1 + \tan(\frac{\pi}{4})} = \frac{1 - 2 \sin^2(\frac{\pi}{4})}{1 + 2 \sin(\frac{\pi}{4}) \cos(\frac{\pi}{4})} \] \[ \frac{1 - 1}{1 + 1} = \frac{1 - 1}{1 + 1} \Rightarrow 0 = 0 \quad \text{(True)} \] ### Conclusion The solutions to the equation are \(x = 0\) and \(x = \frac{\pi}{4}\). If you need further assistance or a different approach, please let me know!