# Prove or disprove $\frac{\sqrt{1+\tan x}}{\cot x} = \frac{1+\sin x}{\cos x}$

I have tried to prove the identity $$\frac{\sqrt{1+\tan x}}{\cot x} = \frac{1+\sin x}{\cos x}$$ by $t$-substitution but seem not to work. Please don't solve(dont 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.

• Counterexample: Take $x=\frac{\pi}{4}$. LHS gives $\sqrt2$ while RHS gives $\sqrt2+1$ Jun 24, 2015 at 18:28
• I guess, this identity makes more sense. Jun 24, 2015 at 18:35
• 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 dont prove identities using angles. Thank you very much for your comments. Jun 24, 2015 at 18:40
• You can't prove an identity by plugging in one angle, but you can certainly disprove a proposed identity.
– Joel
Jun 24, 2015 at 18:53

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.

• How do you get $\frac{\sqrt2+1}{\sqrt2}$ ? Jun 24, 2015 at 18:32
• Typo. I fixed it. @prasunbiswas
– Joel
Jun 24, 2015 at 18:33

As stated, equality is false. Letting $x=0$, the left hand side is $0$ while the right hand side is $1$.

• $x=0$ would make $\cot(x)$ undefined. Jun 24, 2015 at 18:28
• @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 $x=0.00001$. Jun 24, 2015 at 18:29