tan^4 x + tan^2 x+ 1 = 1 - (Sin^2 x)(Cos^2 x)

Please someone help me solve this one so I can get some sleep tonight! I've been trying for 3 days off and on. Pretty please!

Carol

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- Dec 11th 2008, 11:52 AMCarolChangProve tan^4 + tan^2 + 1 = 1 - Sin^2 Cos^2
tan^4 x + tan^2 x+ 1 = 1 - (Sin^2 x)(Cos^2 x)

Please someone help me solve this one so I can get some sleep tonight! I've been trying for 3 days off and on. Pretty please!

Carol - Dec 11th 2008, 12:02 PMChop Suey
This identity is false. Are you sure you copied it correctly?

- Dec 11th 2008, 01:06 PMCarolChang
tan^4Ө + tan^2Ө + 1 = 1 + (sin^2Ө)(cos^2Ө)

Here it is, pasted directly from the homework. If it's false, I will be relieved. Is there a way to prove that it's false?

Carol - Dec 14th 2008, 07:58 PMScopur
Working on the left hand side...

$\displaystyle tan^2x + tan^4x +1$

$\displaystyle = \frac{1}{cos^2x} + tan^4x $ ... the entire left hand side can be manipulated using this.. $\displaystyle 1 +tan^2x= sec^2 x$

$\displaystyle = \frac{1}{cos^2x} + \left(\frac{1}{cos^2x} -1 \right) \left(\frac{1}{cos^2x} -1 \right) $

$\displaystyle = \frac{1}{cos^2x} + \left(\frac{1}{cos^2x} -1 \right) \left(\frac{1}{cos^2x} -1 \right) $

$\displaystyle = \frac{1}{cos^4x} - \frac{1}{cos^2x} +1 $

$\displaystyle = \frac{1}{cos^4x} - tan^2x $

$\displaystyle = 1 $

Is the RHS = 1?