Prove that the following identity is true. sin4 t - cos4 t / sin2 t cos2 t = sec2 t - csc2 t The equation shown is:\[\frac{\sin^4 t - \cos^4 t}{\sin^2 t \cos^2 t} = \sec^2 t - \csc^2 t\]### Explanation:This mathematical expression involves trigonometric identities:1. **Numerator:** \(\sin^4 t - \cos^4 t\) is a difference of squares, which can be factored as \((\sin^2 t + \cos^2 t)(\sin^2 t - \cos^2 t)\). Using the Pythagorean identity, \(\sin^2 t + \cos^2 t = 1\), the numerator simplifies to \(\sin^2 t - \cos^2 t\).2. **Denominator:** \(\sin^2 t \cos^2 t\) is a product of squares of the sine and cosine functions.3. **Right Side of the Equation:** The expression \(\sec^2 t - \csc^2 t\) involves the secant and cosecant functions, which are reciprocals of cosine and sine, respectively. The identity can further be explored using \(\sec^2 t = 1/\cos^2 t\) and \(\csc^2 t = 1/\sin^2 t\).### Educational Purpose: This equation showcases the application of trigonometric identities and algebraic manipulation in simplifying and equating trigonometric expressions. It's a great example for exploring the connections between different trigonometric functions and identities.

Given that sin4t-cos4tsin2tcos2t=sec2t-cosec2tTo prove the identity sin4t-cos4tsin2tcos2t=sec2t-cosec2t. Consider first RHS=sec2t-cosec2t Now sec2t-cosec2t can be written as, sec2t-cosec2t=1cost2-1sint2 After simplification, sec2t-cosec2t=sin2t-cos2tcos2tsin2t We know the identity sin2t+cos2t=1 Multiply both sides of sin2t+cos2t=1 by sin2t, sin4t+cos2tsin2t=sin2t, Substitute in sec2t-cosec2t=sin2t-cos2tcos2tsin2t, sec2t-cosec2t=sin4t+cos2tsin2t-cos2tcos2tsin2t Now substitute sin2t=1-cos2t in sec2t-cosec2t=sin4t+cos2tsin2t-cos2tcos2tsin2t , sec2t-cosec2t=sin4t+cos2t(1-cos2t)-cos2tcos2tsin2t =sin4t+cos2t-cos4t-cos2tcos2tsin2t =sin4t-cos4tcos2tsin2t Which is equal to LHS. Therefore sin4t-cos4tsin2tcos2t=sec2t-cosec2t.