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Verify: $$\sec^2x + \tan^2x = (1-\sin^4x)\sec^4x$$

My solution:

$$ \begin{align}\sec^2x+\tan^2x&=\frac{1}{\cos^2x}+\frac{\sin^2x}{\cos^2x}\\ &=\frac{1+\sin^2x}{\cos^2x}\\ &=\frac{1+\sin^2x}{\cos^2x}\cdot\frac{1-\sin^2x}{1-\sin^2x}\\ &=\frac{1-\sin^4x}{\cos^2x\cdot\cos^2x}\\ &=\frac{1-\sin^4x}{\cos^4x}\\ &=\frac{1}{\cos^4x}-\frac{\sin^4x}{\cos^4x}\\ &=\sec^4x-\sin^4x\sec^4x\\ &=\sec^4x(1-\sin^4x)\\ \end{align}$$

Is it incorrect to multiply in $1-\sin^2x$ like in the fourth equality?

3 Answers3

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It's okay because $1-\sin^2 x=\cos^2 x$ so it can only become $0$ if $\cos x$ is zero, wich can be excluded due to the domain of the equality, thus no division by zero can occur and you may proceed.

AlexR
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  • So what your saying is if I'm multiplying in other identities I only have to worry about whether the initial domains hold throughout the entire verification process? –  Nov 19 '14 at 11:23
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    @Earth52 Yes. You need to check that whenever you divide by something, that something cannot become zero in all of the original domain. Here $\frac{1-\sin^2 x}{1-\sin^2 x}=(1-\sin^2 x) \underline{\div (1-\sin^2 x)}$ so it falls into that case. – AlexR Nov 19 '14 at 11:34
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No it's absolutely correct

You may be worried about What if (Sin^2)x= 1
But it can not be achieved in the domain of the equality otherwise (tan^2)x and (sec^2)x at (sin^2)x=1 will not be defined

DSinghvi
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we have to show that $\sec(x)^2+\tan(x)^2-(1-\sin(x)^4)\sec(x)^4=0$, the left hand side is equivalent to $\frac{\sin(x)^4-1}{\cos(x)^4}+\frac{\sin(x)^2+1}{\cos(x)^2}=\frac{(\sin(x)^2-1)(\sin(x)^2+1)+(\sin(x)^2+1)\cos(x)^2}{\cos(x)^4}$
the numerator is equivalent to
$(\sin(x)^2+1)(\sin(x)^2+\cos(x)^2-1)=0$ since $\sin(x)^2+\cos(x)^2=1$

Dr. Sonnhard Graubner
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    You are so off with your answers, in terms of whether they address (or fail to address) the question posted! Please take more time to understand the actual questions that are posted (before jumping in to answer). Then, take a little more time composing answers that actually answer (well) the given question. – amWhy Nov 19 '14 at 11:49
  • no i'm not off, my proof shows that we don't need to multiply with $1-\sin(x)^2$ which can be zero – Dr. Sonnhard Graubner Nov 19 '14 at 11:51
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    I did not say you were wrong. I just said that you seem to have a habit of posting answers which have little to do with the posted question. You also need to start adding a bit more prose to actually explain what/why you are doing. [See AlexR's post to see why there is no problem, in this case, when multiplying by $1-\sin^2 x = \cos^2 x$] – amWhy Nov 19 '14 at 11:54
  • @Dr.SonnhardGraubner You do implicitly multiply by $\cos^2 = 1-\sin^2 x$ when expanding the fractions. – AlexR Nov 19 '14 at 12:03