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