g) \( \sec ^{3} x=\frac{\sec ^{3} x-\sec x \tan ^{2} x}{(1-\operatorname{sen} x)(1+\operatorname{sen} x)} \)

To verify the trigonometric identity: \[ \sec^3 x = \frac{\sec^3 x - \sec x \tan^2 x}{(1 - \sin x)(1 + \sin x)} \] let's simplify the right-hand side (RHS) step by step and show that it equals the left-hand side (LHS). ### Step 1: Simplify the Denominator The denominator of the RHS is: \[ (1 - \sin x)(1 + \sin x) \] This is a difference of squares, which simplifies to: \[ (1 - \sin x)(1 + \sin x) = 1 - \sin^2 x \] Using the Pythagorean identity \(1 - \sin^2 x = \cos^2 x\), we get: \[ 1 - \sin^2 x = \cos^2 x \] ### Step 2: Rewrite the RHS with the Simplified Denominator Substitute the simplified denominator back into the RHS: \[ \frac{\sec^3 x - \sec x \tan^2 x}{\cos^2 x} \] ### Step 3: Simplify the Numerator Recall the definitions of the trigonometric functions: \[ \sec x = \frac{1}{\cos x}, \quad \tan x = \frac{\sin x}{\cos x} \] First, express \(\tan^2 x\) in terms of \(\sin^2 x\) and \(\cos^2 x\): \[ \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \] Now, substitute \(\sec x\) and \(\tan^2 x\) into the numerator: \[ \sec^3 x - \sec x \tan^2 x = \frac{1}{\cos^3 x} - \frac{1}{\cos x} \cdot \frac{\sin^2 x}{\cos^2 x} \] Simplify the expression: \[ \frac{1}{\cos^3 x} - \frac{\sin^2 x}{\cos^3 x} = \frac{1 - \sin^2 x}{\cos^3 x} \] Again, using the Pythagorean identity \(1 - \sin^2 x = \cos^2 x\): \[ \frac{1 - \sin^2 x}{\cos^3 x} = \frac{\cos^2 x}{\cos^3 x} = \frac{1}{\cos x} = \sec x \] ### Step 4: Combine the Simplified Numerator and Denominator Now, the RHS becomes: \[ \frac{\sec x}{\cos^2 x} \] Express \(\sec x\) in terms of \(\cos x\): \[ \frac{\frac{1}{\cos x}}{\cos^2 x} = \frac{1}{\cos^3 x} = \sec^3 x \] ### Conclusion After simplifying, both sides of the equation are equal: \[ \sec^3 x = \sec^3 x \] Thus, the given identity is **verified**.