ou need to prove the following trigonometric identity such that:

tan^2q cos^2q= (sec^2q-1)(1-sin^4q)/(1 +sin^2q)

Since denominator never cancels, for any value of q, `(1 +sin^2q)!= 0` , hence, there are no restrictions to the given expression.

You need to remember that `tan q = sin q/cos q` , hence, you may substitute...

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ou need to prove the following trigonometric identity such that:

tan^2q cos^2q= (sec^2q-1)(1-sin^4q)/(1 +sin^2q)

Since denominator never cancels, for any value of q, `(1 +sin^2q)!= 0` , hence, there are no restrictions to the given expression.

You need to remember that `tan q = sin q/cos q` , hence, you may substitute `(sin q/cos q)^2` for `tan^2 q` such that:

`(sin q/cos q)^2cos^2q = (sec^2q-1)(1-sin^4q)/(1 +sin^2q) `

Reducing by `cos^2 q` to the left side yields:

`sin^2 q = (sec^2q-1)(1-sin^4q)/(1 +sin^2q) `

Performing the cross multiplication yields:

`sin^2 q(1 + sin^2 q) = (sec^2q-1)(1-sin^4q)`

You need to remember that `sec q = 1/cos q` , hence, you should substitute `1/(cos^2 q)` for `sec^2 q` such that:

`sin^2 q(1 + sin^2 q) = (1/(cos^2 q) - 1)(1 - sin^4q)`

Bringing the terms to the right side to a common denominator yields:

`sin^2 q(1 + sin^2 q) = (1-(cos^2 q))/(cos^2 q)(1 - sin^4q)`

You need to remember that `1- cos^2 q = sin^2 q` , hence, you may substitute `sin^2 q ` for `1 - cos^2 q` to the right side such that:

`sin^2 q(1 + sin^2 q) = ((sin^2 q)/(cos^2 q))(1 - sin^4q)`

You may convert the difference of squares `1 - sin^4q` into a product such that:

`1 - sin^4q = (1 - sin^2 q)(1 + sin^2 q)`

`sin^2 q(1 + sin^2 q) = ((sin^2 q)/(cos^2 q))((1 - sin^2 q)(1 + sin^2 q))`

Substituting `cos^2 q` `for 1 - sin^2 q` , to the right side, yields:

`sin^2 q(1 + sin^2 q) = ((sin^2 q)/(cos^2 q))(cos^2 q)(1 + sin^2 q)`

Reducing by `cos^2 q ` to the right yields:

`(sin^2 q)(1 + sin^2 q) = (sin^2 q)(1 + sin^2 q)`

**Hence, performing the substitutions above, yields both sides equal, hence, the given identity `tan^2q cos^2q= (sec^2q-1)(1-sin^4q)/(1 +sin^2q)` holds.**