Prove tan^4(w) + 2tan^2(w) +1 = sec^4(w)

[mgiddy911]

i need help showing the folowing is valid:

tan^4(w) + 2tan^2(w) +1 = sec^4(w)

I am pretty lost, I know I should expand the left side and go from there. SO far I have gotten:

[sin^4(w)/cos^4(w)] + [2sin^2(w)/cos^2(w)] + [sin^2(w) +cos^2(w)]

and then I am stuck when I try and add these terms together, Imay have made mistakes some where along the line in adding the fractions, could someone point me in the right direction? am I right so far?
the farthest i have gotten on the left side is:
[sin^4(w) + 2sin^2(w)cos^2(w) + sin^2(w)cos^4(w) +cos^6(w)] /cos^4(w)

 

[Curious3141]

i need help showing the folowing is valid:

tan^4(w) + 2tan^2(w) +1 = sec^4(w)

I am pretty lost, I know I should expand the left side and go from there. SO far I have gotten:

[sin^4(w)/cos^4(w)] + [2sin^2(w)/cos^2(w)] + [sin^2(w) +cos^2(w)]

and then I am stuck when I try and add these terms together, Imay have made mistakes some where along the line in adding the fractions, could someone point me in the right direction? am I right so far?
the farthest i have gotten on the left side is:
[sin^4(w) + 2sin^2(w)cos^2(w) + sin^2(w)cos^4(w) +cos^6(w)] /cos^4(w)

Observe that the LHS is just {(1 + \tan^2{w})}^2 = {(\sec^2{w})}^2 = \sec^4{w}

If you're not allowed to assume that 1 + \tan^2{w} = \sec^2{w}, just divide \sin^2{w} + \cos^2{w} = 1 throughout by \cos^2{w} and see what you get.

 

[maverick280857]

Another way out if you're not "allowed" to do some other things is to subtract 1 from each side of the equality to be proved and use a variant of the identity given by Curious3141.

Note that what you have to prove is not strictly identity for all w, unless you bend the definition of an identity to accommodate it. The terms on the left hand side are not defined when x = (2n+1)\frac{\pi}{2} where n is an integer. Same goes for the term on the right hand side. However, it would be better to say that the terms tend to \infty as x approaches this value from the left or right and hence, this minor argument does not really matter much.

Cheers
Vivek