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.

Asked on by rosemin

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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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