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- Jan 26, 2012

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Thread starter
- Admin
- #1

- Jan 26, 2012

- 4,055

--------------------

Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter
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- #2

- Jan 26, 2012

- 4,055

Congratulations to the following members for their correct solutions:

1) MarkFL

2) Sudharaka

3) anemone

4) BAdhi

5) Petrus

6) kaliprasad

7) agentmulder

8) Reckoner

Solution (from Reckoner):

By factoring the original left-hand side, we have

\begin{align*}

\sec^4 x - \tan^4 x &= \left(\sec^2 x + \tan^2 x\right)\left(\sec^2 x - \tan^2 x\right)\\

&= \left(\sec^2 x + \tan^2 x\right)(1)\\

&= \left(\sec^2 x + \tan^2 x\right).

\end{align*}

Solution (from BAdhi):

\text{L.H.S.}&= \sec^4x-\tan^4x\\

&=\left[\sec^2x\right]^2-\tan^4x\\

&=\left[1+\tan^2x\right]^2-\tan^4x\\

&=1+2\tan^2x+\tan^4x-\tan^4x\\

&=1+2\tan^2x\\

&=(\sec^2x-\tan^2x)+2\tan^2x\\

&=\sec^2x+\tan^2x

&=\text{R.H.S.}

\end{align*}$$

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