Prove the following identity: (tan^4)t + (tan^2)t + 1 = [(1 - (sin^2)t * (cos^2) t] / (cos^4) t I asked this same question before but the answer was not very clear. Could you please explain it...
Prove the following identity:
(tan^4)t + (tan^2)t + 1 = [(1 - (sin^2)t * (cos^2) t] / (cos^4) t
I asked this same question before but the answer was not very clear. Could you please explain it again.
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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 +...
(The entire section contains 193 words.)
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Note: We'll put the argument of the function, in this case t, near the function, and the power the function is raised, outside the brackets.
We'll re-write the right side expression in this way:
1/(cos t)^4 - (sin t)^2*(cos t)^2/(cos t)^4
We'll simplify by (cos t)^2:
1/(cos t)^4 - (sin t)^2/(cos t)^2
We know that the tangent function could be written as:
tan t = sin t/cos t
We'll raise to square:
(tan t)^2 = (sin t)^2/(cos t)^2
We also know that 1/(cos t)^2 = 1 + (tan t)^2 (fundamental formula of trigonometry)
We'll raise to square:
[1/(cos t)^2]^2 = [1 + (tan t)^2]^2
1/(cos t)^4 = 1 + (tan t)^4 + 2(tan t)^2
The right side expression will become:
1/(cos t)^4 - (sin t)^2/(cos t)^2 = 1 + (tan t)^4 + 2(tan t)^2 - (tan t)^2
We'll combine like terms:
1/(cos t)^4 - (sin t)^2/(cos t)^2 = 1 + (tan t)^4 + (tan t)^2
We notice that we've get exactly the left side expression:
(tan t)^4 + (tan t)^2 + 1 = 1 + (tan t)^4 + (tan t)^2 q.e.d.
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