prove using identities

[btrfly24]

here's the problem:

prove that tan^4(x) = sec^2(x)-csc^2(x) + csc^2(x)cos^2(x) all divided by cot^2(x)

Now I have the whole answer written out in steps, but I can't figure out how they even got the first step of:

sec^2(x) - csc^2(x)(1-cos^2(x)) divided by cot^2(x)

I know that the 1-cos^2(x) = sin^2(x) but how did that come from the original equation? Please help, and thanks in advance

Catherine.

 

[galactus]

\(\displaystyle \L\\\frac{sec^{2}(x)-csc^{2}(xx)+csc^{2}(x)cos^{2}(x)}{cot^{2}(x)}\)

Now, expand it out so you can see it better and use identities:

\(\displaystyle \L\\\underbrace{\frac{sec^{2}(x)}{cot^{2}(x)}}_{\text{sec^2(x)tan^2(x)}}-\overbrace{\frac{csc^{2}(x)}{cot^{2}(x)}}^{\text{sec^2(x)}}+\underbrace{\frac{csc^{2}(x)cos^{2}(x)}{cot^{2}(x)}}_{\text{1}}\)

So, you have:

\(\displaystyle \L\\sec^{2}(x)tan^{2}(x)+\underbrace{1-sec^{2}(x)}_{\text{-tan^2(x))\)

Then you have:

\(\displaystyle \L\\sec^{2}(x)tan^{2}(x)-tan^{2}(x)\)

Factor out \(\displaystyle tan^{2}(x)\)

\(\displaystyle \L\\tan^{2}(x)\underbrace{(sec^{2}(x)-1)}_{\text{tan^2(x)}}\)

Now you have:

\(\displaystyle \H\\tan^{2}(x)tan^{2}(x)=tan^{4}(x)\)

 

[soroban]

Hello, Catherine!

Prove that: \(\displaystyle \L\:\tan^4(x) \:= \:\frac{\sec^2(x)\,-\,csc^2(x)\,+\,\csc^2(x)\cos^2(x)}{\cot^2(x)}\)

I can't figure out how they even got the first step of:
. . \(\displaystyle \L\frac{\sec^2(x)\,-\,\csc^2(x)(1\,-\,\cos^2(x)}{\cot^2(x)}\)

The numerator is: \(\displaystyle \:\sec^2(x)\,-\,\csc^2(x)\,+\,\csc^2(x)\cos^2(x)\)


They factored out -\(\displaystyle \csc^2(x)\) from the last two terms:

. . \(\displaystyle \sec^2(x)\,-\,\csc^2x(x)\cdot\underbrace{\left[1\,-\,\cos^2(x)\right]}\)
. . \(\displaystyle =\;\sec^2(x)\,-\,\underbrace{\csc^2x(x)\cdot\sin^2(x)}\)
. . . . .\(\displaystyle =\;\underbrace{\sec^2(x)\;\;-\;\;1}\)
. . . . . . . . \(\displaystyle = \;\tan^2(x)\)

The problem becomes: \(\displaystyle \L\:\frac{\tan^2(x)}{\cot^2(x)} \;=\;\frac{\tan^2(x)}{\frac{1}{\tan^2(x)}} \;=\;\tan^4(x)\)