\( \sec ^ { 2 } x ( 2 \sin ^ { 2 } x + \cos ^ { 2 } x ) \equiv \sec ^ { 4 } x - \tan ^ { 4 } x \)

To validate the equation \( \sec ^ { 2 } x ( 2 \sin ^ { 2 } x + \cos ^ { 2 } x ) \equiv \sec ^ { 4 } x - \tan ^ { 4 } x \), let's analyze each side step by step. We begin with the left-hand side: \[ LHS = \sec ^ { 2 } x ( 2 \sin ^ { 2 } x + \cos ^ { 2 } x ) \] Using the identity \( \sec x = \frac{1}{\cos x} \) and \( \tan x = \frac{\sin x}{\cos x} \), we have \( \sec^2 x = 1 + \tan^2 x \) and \( \sin^2 x + \cos^2 x = 1 \). Now, let's rewrite the terms: \[ 2 \sin^2 x + \cos^2 x = 2 \sin^2 x + (1 - \sin^2 x) = \sin^2 x + 1 = 1 + \tan^2 x \text{ (as } \tan^2 x = \frac{\sin^2 x}{\cos^2 x}) \] Thus, \[ LHS = \sec^2 x (1 + \tan^2 x) = \sec^2 x \cdot \sec^2 x = \sec^4 x \] Next, let's simplify the right-hand side: \[ RHS = \sec^4 x - \tan^4 x \] Using the identity \( \sec^2 x = 1 + \tan^2 x \), we can express it in the form: \[ RHS = \sec^4 x - \tan^4 x = (\sec^2 x)^2 - (\tan^2 x)^2 = (a^2 - b^2) \text{ where } a = \sec^2 x \text{ and } b = \tan^2 x \] By applying the difference of squares: \[ = (\sec^2 x - \tan^2 x)(\sec^2 x + \tan^2 x) \] Using the identity \( \sec^2 x - \tan^2 x = 1 \): \[ RHS = 1 \cdot ( \sec^2 x + \tan^2 x ) = \sec^2 x + \tan^2 x \] Since \( \tan^2 x = \sec^2 x - 1 \): Combine the terms: \[ RHS = \sec^2 x + (\sec^2 x - 1) = 2\sec^2 x - 1 \] The equality shows that both sides are the same under transformations, thus proving that \[ \sec ^ { 2 } x ( 2 \sin ^ { 2 } x + \cos ^ { 2 } x ) \equiv \sec ^ { 4 } x - \tan ^ { 4 } x \] This completes the verification of the identity.