\( \left. \begin{array} { l } { 3 \left. \begin{array} { l } { \frac { 1 + \tan ^ { 2 } x } { 1 - \tan ^ { 2 } x } = \frac { 1 } { 2 \cos ^ { 2 } x - 1 } } \\ { 1 + \sin x \quad 1 - \sin x } \\ { } \end{array} \right. = 4 } \end{array} \right. \)

To solve the given problem, we need to analyze the equation step by step. The equation provided is: \[ \frac{1 + \tan^2 x}{1 - \tan^2 x} = \frac{1}{2 \cos^2 x - 1} \] We also have the expression \(1 + \sin x\) and \(1 - \sin x\) which seems to be part of a larger context, but let's focus on solving the equation first. ### Step 1: Simplify the Left Side Using the identity \(1 + \tan^2 x = \sec^2 x\), we can rewrite the left side: \[ \frac{\sec^2 x}{1 - \tan^2 x} \] Next, we can express \(1 - \tan^2 x\) in terms of sine and cosine: \[ 1 - \tan^2 x = 1 - \frac{\sin^2 x}{\cos^2 x} = \frac{\cos^2 x - \sin^2 x}{\cos^2 x} = \frac{\cos 2x}{\cos^2 x} \] Thus, the left side becomes: \[ \frac{\sec^2 x}{\frac{\cos 2x}{\cos^2 x}} = \frac{\sec^2 x \cdot \cos^2 x}{\cos 2x} = \frac{1}{\cos 2x} \] ### Step 2: Simplify the Right Side The right side is already in a simplified form: \[ \frac{1}{2 \cos^2 x - 1} \] ### Step 3: Set the Two Sides Equal Now we set the two sides equal to each other: \[ \frac{1}{\cos 2x} = \frac{1}{2 \cos^2 x - 1} \] ### Step 4: Cross Multiply Cross multiplying gives us: \[ 2 \cos^2 x - 1 = \cos 2x \] Using the double angle identity for cosine, we know that: \[ \cos 2x = 2 \cos^2 x - 1 \] This means that both sides of the equation are equal, confirming that the equation holds true for all \(x\). ### Step 5: Analyze the Additional Expression The additional expression \(1 + \sin x\) and \(1 - \sin x\) does not seem to directly relate to the equation we solved. If it is part of a larger context or if there are specific values or conditions to apply, please provide that information for further analysis. ### Conclusion The equation \(\frac{1 + \tan^2 x}{1 - \tan^2 x} = \frac{1}{2 \cos^2 x - 1}\) holds true for all \(x\) where the expressions are defined. If you have any specific values or further context regarding \(1 + \sin x\) and \(1 - \sin x\), please let me know!