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Going steadily through my book, I found this exercise to resolve

$$ \sec{x}\left(\sin^3x + \sin x \cos^2x\right)=\tan{x}$$

Here's how I resolve it ($LHS$) and again bear with me as I truly reverting to a feeling of vulnerability, like a child actually

As $\sec x$ is equal to $\frac{1}{\cos x}$

That leads us to this

$$\frac{(\sin^3x+\sin x\cos^2x)}{\cos x}$$

I'm factorizing one $\sin x$

$$\frac{\sin x(\sin^2x+\cos^2x)}{\cos x} = \frac{\sin x(1)}{\cos x} = \tan x$$

That seems to work otherwise I completly messed this up

Reading the book's solution, I have something different...

$$\begin{align*} LHS&=\frac{\sin^3x}{\cos x}+ \sin x \cos x \\[4pt] &=\frac{\sin x}{\cos x}-\frac{\sin x\cos^2x}{\cos x}+\sin x\cos x\\[4pt] &= \tan x\end{align*}$$

What did I miss?

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    $\begingroup$ They messed this up. $\endgroup$ – Yves Daoust May 13 '16 at 10:00
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    $\begingroup$ Shorter: multiplying both members by $\cot(x)$ yields $\sin^2(x)+\cos^2(x)=1$. $\endgroup$ – Yves Daoust May 13 '16 at 10:04
  • $\begingroup$ Hi @YvesDaoust you see this is where dogma starts. My way is better than your way and at the end, you are disgusting a generation of people but for what purpose...? $\endgroup$ – Andy K May 13 '16 at 10:15
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    $\begingroup$ There is absolutely no intent to disgust anyone. You seem to miss that my first remark was an compliment to you. There is beauty in making things short. $\endgroup$ – Yves Daoust May 13 '16 at 10:22
  • $\begingroup$ @YvesDaoust I understood correctly what you said. I was more empathetic about these 2 chaps who went up to create a complicate solution for what it was. They could have done something simplier and also mentionned that, there were other ways (pluralist views) to solve that problem. As we said J'avais bien compris :) $\endgroup$ – Andy K May 13 '16 at 10:24
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So the book does about the same thing as you but in a different order (your solution is hence correct too). By the trigonometric one we get $\sin^2 x= (1-\cos^2 x) $. Thus $$\frac{\sin^3 x}{\cos x}=\frac{\sin x (\sin^2 x)}{\cos x}= \frac{\sin x(1-\cos^2 x)}{\cos x} = \frac{\sin x-\sin x\cos^2 x}{\cos x}= \frac{\sin x}{\cos x} -\frac{\sin x\cos ^2 x}{\cos x}$$ Thus $$LHS =\frac{\sin^3 x}{\cos x}+\sin x\cos x= \frac{\sin x}{\cos x} -\frac{\sin x\cos ^2 x}{\cos x} +\sin x\cos x = \frac{\sin x}{\cos x}=\tan x$$

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  • $\begingroup$ Hi @ove-ahlman thanks for your answer. I'll write the two authors an email once I've finished their book. it is an interesting one but some of the demos are too cumbersome, for what it is $\endgroup$ – Andy K May 13 '16 at 10:29
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You missed nothing. The book is doing the same you did, only in a more cumbersome way.

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  • $\begingroup$ Hi @martin-argerami, many thanks for your answer. I wish to give you the points but well, Ove went on a lengthy explanation to explain these 2 chaps way of thoughts. Thanks for your answer :) $\endgroup$ – Andy K May 13 '16 at 10:31

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