We have to prove the identity: (tan^4)t + (tan^2)t + 1 = [(1 - (sin^2)t * (cos^2) t] / (cos^4) t

First we write it in a more regular way:

(tan t)^4 + (tan t)^2 + 1 = [(1 - (sin t)^2 * (cos t)^2] / (cos t)^4

Let's start with the left hand side:

(tan t)^4 + (tan t)^2 + 1

use tan t = sin t / cos t

=> (sin t / cos t)^4 + (sin t / cos t)^2 + 1

=> (sin t)^4 / (cos t)^4 + (sin t)^2 / (cos t)^2 + 1

make the denominator the same for all the terms

=> (sin t)^4/(cos t)^4 + (sin t)^2*(cos t)^2/(cos t)^4 + (cos t)^4/ (cos t)^4

=> [(sin t)^4 + (sin t)^2*(cos t)^2 + (cos t)^4]/ (cos t)^4

Now we know that(a +b)^2 = a^2 + b^2 + 2ab

=> a^2 + b^2 + ab = (a + b)^2 - ab

take (sin t)^2 = a and (cos t)^2 = b

[(sin t)^4 + (sin t)^2*(cos t)^2 + (cos t)^4]/ (cos t)^4

=> [((sin t)^2 + (cos t)^2)^2 - (sin t)*(cos t)]/( cos t)^4

now use the relation (sin t)^2 + (cos t)^2 = 1

=> [1 - (sin t)*(cos t)]/( cos t)^4

which is the right hand side.

**Therefore we prove the identity: (tan t)^4 + (tan t)^2 + 1 = [(1 - (sin t)^2 * (cos t)^2] / (cos t)^4**