Answer
$(1-\sin^{2}t+\cos^{2}t)^{2}+4\sin^{2}t\cos^{2}t=4\cos^{2}t$
Work Step by Step
$(1-\sin^{2}t+\cos^{2}t)^{2}+4\sin^{2}t\cos^{2}t=4\cos^{2}t$
On the left side of the equation, substitute $1$ with $\sin^{2}t+\cos^{2}t$ and simplify:
$(\sin^{2}t+\cos^{2}t-\sin^{2}t+\cos^{2}t)^{2}+4\sin^{2}t\cos^{2}t=4\cos^{2}t$
$(2\cos^{2}t)^{2}+4\sin^{2}t\cos^{2}t=4\cos^{2}t$
$4\cos^{4}t+4\sin^{2}t\cos^{2}t=4\cos^{2}t$
On the left side of the equation, take out common factor $4\cos^{2}t$:
$4\cos^{2}t(\cos^{2}t+\sin^{2}t)=4\cos^{2}t$
Since $\cos^{2}t+\sin^{2}t=1$, the identity is proved:
$4\cos^{2}t=4\cos^{2}t$
