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I have tried to prove the identity \begin{equation} \frac{\sqrt{1+\tan x}}{\cot x} = \frac{1+\sin x}{\cos x} \end{equation} by $t$-substitution but seem not to work. Please don't solve(don`t post the answer on this site) this question for me just try it and give me hints on how I should go about it. Or if you like you can post but I wanted to try using some hints that you would give first.

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    $\begingroup$ Counterexample: Take $x=\frac{\pi}{4}$. LHS gives $\sqrt2$ while RHS gives $\sqrt2+1$ $\endgroup$ Jun 24, 2015 at 18:28
  • $\begingroup$ I guess, this identity makes more sense. $\endgroup$ Jun 24, 2015 at 18:35
  • $\begingroup$ I was argued against in class because initialy I plugged in pi/4 but our tutor proved some incompetence to me and competence to the class and he said you don`t prove identities using angles. Thank you very much for your comments. $\endgroup$ Jun 24, 2015 at 18:40
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    $\begingroup$ You can't prove an identity by plugging in one angle, but you can certainly disprove a proposed identity. $\endgroup$
    – Joel
    Jun 24, 2015 at 18:53

2 Answers 2

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If $x=\pi/4$ we have: $$\frac{\sqrt{1+1}}{1}=\sqrt{2}$$ on the left, but $$\frac{1+ \frac{1}{\sqrt{2}}}{1/\sqrt{2}}=\sqrt{2}+1$$ on the right.

These are not equivalent.

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    $\begingroup$ How do you get $\frac{\sqrt2+1}{\sqrt2}$ ? $\endgroup$ Jun 24, 2015 at 18:32
  • $\begingroup$ Typo. I fixed it. @prasunbiswas $\endgroup$
    – Joel
    Jun 24, 2015 at 18:33
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As stated, equality is false. Letting $x=0$, the left hand side is $0$ while the right hand side is $1$.

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    $\begingroup$ $x=0$ would make $\cot(x)$ undefined. $\endgroup$ Jun 24, 2015 at 18:28
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    $\begingroup$ @PrasunBiswas: The function is analytic around $0$ and has a removable singularity - note we may write it as $$\frac{\sin(x)\sqrt{1+\tan(x)}}{\cos(x)}.$$ If you strongly dislike removable singularities, then take $x=0.00001$. $\endgroup$ Jun 24, 2015 at 18:29

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