# 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 $\sqrt2+1$ – Prasun Biswas Jun 24 '15 at 18:28
• I guess, this identity makes more sense. – Dietrich Burde Jun 24 '15 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. – DOCTOR NGILAZI BANDA JOSHUA Jun 24 '15 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 '15 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.
• How do you get $\frac{\sqrt2+1}{\sqrt2}$ ? – Prasun Biswas Jun 24 '15 at 18:32
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. – Prasun Biswas Jun 24 '15 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 $x=0.00001$. – Eric Naslund Jun 24 '15 at 18:29