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I know that my solution is wrong because for $x = 0^\circ,$ the results are not the same. Where did I go wrong?

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gerard hotaro
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    Your work is correct. It's just that the given expression is undefined for $x=2k\pi$ and $x=\pi/2+k\pi, k\in\mathbb{Z}$ so you can't plug in $x=0^{\circ}$ or $x=90^{\circ}$. Try plugging in anything else for which the expression is defined, for instance $x=30^{\circ}$. – bjorn93 Dec 07 '21 at 15:20
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    Your first step falsely assumes that $x$ cannot be $180^\circ;$ your final step drops the implicit condition that $x$ cannot be $0^\circ.$ – ryang Dec 07 '21 at 17:40

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  1. The given expression $$\frac{\tan x+\sec x-1}{\tan x-\sec x +1}\tag L$$ has the implicit condition $$\tan x-\sec x+1\ne0\quad\text{AND}\quad\cos x\ne0\\ \cos x-\sin x\ne1\quad\text{AND}\quad \cos x\ne0\\ \sqrt2\cos\left(x+\frac\pi4\right)\ne1\quad\text{AND}\quad x\ne2n\pi\pm\frac\pi2\\ x\ne2n\pi\:\text{ or }\:(2n\pm1) \frac\pi2,$$ whereas your new expression $$\tan x+\sec x\tag R$$ has the implicit condition $$\cos x\ne0\\ x\ne(2n\pm1) \frac\pi2.$$

    In fact, the identity $$\frac{\tan x+\sec x-1}{\tan x-\sec x +1}\equiv\tan x+\sec x$$ has domain $$\mathbb R{\setminus}\left\{x:\exists n{\in}\mathbb Z\:\:x= 2n\pi\:\text{ or }\:(2n\pm1)\frac\pi2\right\}.$$

  2. In your first step, you carelessly introduced the restriction $$\tan x+\sec x+1\ne0;$$ then in the final step, you invalidly discarded the restriction $$\tan x\ne0;$$ these missteps in your simplification ultimately expanded the domain by adding the even multiples of $\pi$ to it.

ryang
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