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I've got this task,

$$\frac{\sin^6x}{1 - \tan^2x} + \frac{\cos^6x}{1-\cot^2x} = - \sin^2x \cdot \cos^2x$$

I've gotten to this part but I'm not even sure it's the correct path to take since if i actually add the two fractions i get even worse situation with nothing to remove.

$$\frac{\sin^6x \cdot \cos^2x}{\cos^2x - \sin^2x} + \frac{\cos^6x \cdot \sin^2x}{\sin^2x - \cos^2x} = - \sin^2x \cdot \cos^2x$$

Thanks

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  • $\begingroup$ What is $ctg^2$? $\endgroup$ – Jan Oct 25 '15 at 16:59
  • $\begingroup$ You have made substantial progress, and are in fact quite close to the end. Bring to a common denominator, that's just a sign change. and now factor the top. $\endgroup$ – André Nicolas Oct 25 '15 at 17:03
  • $\begingroup$ $ctg^2 = 1/tg^2$ $\endgroup$ – simon101 Oct 25 '15 at 17:05
  • $\begingroup$ @AndréNicolas okay will try. Thanks $\endgroup$ – simon101 Oct 25 '15 at 17:05
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Oct 25 '15 at 17:07
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You almost have it: factor out $\sin^2x\cos^2x$ from the numerator \begin{align*} \frac{\sin^6x\cos^2x}{\cos^2x-\sin^2x}&+\frac{\cos^6x\sin^2x}{\sin^2x-\cos^2x}= -\sin^2x\cos^2x\frac{\cos^4x-\sin4x}{\cos^2x-\sin^2x}\\ &=-\sin^2x\cos^2x\frac{\cos^2x-\sin2x}{\cos^2x-\sin^2x}=-\sin^2x\cos^2x. \end{align*}

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  • $\begingroup$ How did you get $cos^4x - sin^4x$ to be $cos^2x - sin^2x$ in step 2->3 $\endgroup$ – simon101 Oct 25 '15 at 17:34
  • $\begingroup$ $A^4-B^4=(A^2-B^2)(A^2+B^2)$, and $A^2+B^2=\dots$ $\endgroup$ – Bernard Oct 25 '15 at 17:36
  • $\begingroup$ You're welcome. Glad I could help! $\endgroup$ – Bernard Oct 25 '15 at 17:49
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$$\frac{\sin^6x \cdot \cos^2x}{\cos^2x - \sin^2x} + \frac{\cos^6x \cdot \sin^2x}{\sin^2x - \cos^2x} = - \sin^2x \cdot \cos^2x$$ Multiply by $\sin^2x-\cos^2x$ to get

$$-\sin^6x\cos^2x+\cos^6x\sin^2x=-\sin^2x\cos^2x(\sin^2x - \cos^2x)$$

Expand

$$-\sin^6x\cos^2x+\cos^6x\sin^2x=-\sin^4x\cos^2x+\sin^2x\cos^4 x.$$

Divide by $\sin^2x\cos^2x$ to get

$$-\sin^4x+\cos^4x=-\sin^2x+\cos^2 x.$$

This leads to

$$(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)=-\sin^2x+\cos^2 x.$$

But since $\sin^2x+\cos^2x = 1$, we have equality. You will have to be careful when dividing though since we cannot divide by zero, so you may have to prove those situations as a special cases.

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    $\begingroup$ Thanks! I uopvoted, but because I made the account recently it might not show. $\endgroup$ – simon101 Oct 25 '15 at 17:43

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