Question

Prove that cot pi/24 = root 2 + root 3 + root 4 + root 6

Answer 1

LHS = cot pi/24 = cot 180/24 = cot 7.5 = 7.59
RHS = root2 + root 3 + root4 + root 6 = 7.59
using log tabkes or calculator

Answer 2

Answer and Explanation:

To prove : $\cot \frac{\pi}{24}=\sqrt{2} +\sqrt{3}+\sqrt{4}+\sqrt{6}$

Proof :

Taking LHS,

$LHS=\cot \frac{\pi}{24}$

$=\frac{\cos(\frac{\pi}{24})}{\sin(\frac{\pi}{24})}$

Multiply and divide by $2\cos(\frac{\pi}{24})$

$=\frac{\cos(\frac{\pi}{24})}{\sin(\frac{\pi}{24})}\times \frac{2\cos(\frac{\pi}{24})}{2\cos(\frac{\pi}{24})}$

$=\frac{2\cos^2(\frac{\pi}{24})}{2\sin(\frac{\pi}{24})\cos(\frac{\pi}{24})}$

Applying formula,

$2\cos^2x=1+\cos (2x)\\2\sin x\cos x=\sin 2x$

$=\frac{1+\cos(\frac{\pi}{12})}{\sin(\frac{\pi}{12})}$

$=\frac{1+\cos(\frac{\pi}{4}-\frac{\pi}{6})}{\sin(\frac{\pi}{4}-\frac{\pi}{6})}$

Applying formula,

$\cos (A-B)=\cos A\cos B+\sin A\sin B\\\sin (A-B)=\sin A\cos B-\cos A\sin B$

$=\frac{1+\cos(\frac{\pi}{4})\cos(\frac{\pi}{6})+\sin(\frac{\pi}{4})\sin(\frac{\pi}{6})}{\sin(\frac{\pi}{4})\cos(\frac{\pi}{6})-\cos(\frac{\pi}{4})\sin(\frac{\pi}{6})}$

Substituting the values,

$=\frac{1+(\frac{1}{\sqrt2})(\frac{\sqrt3}{2})+(\frac{1}{\sqrt2})(\frac{1}{2})}{(\frac{1}{\sqrt2})(\frac{\sqrt3}{2})-(\frac{1}{\sqrt2})(\frac{1}{2})}$

$=\frac{2\sqrt2+\sqrt3+1}{\sqrt3-1}$

Rationalizing by multiply Nr. and Dr. by $\sqrt3+1$

$=\frac{2\sqrt2+\sqrt3+1}{\sqrt3-1}\times \frac{\sqrt3+1}{\sqrt3+1}$

$=\frac{2\sqrt6+3+\sqrt3+2\sqrt2+\sqrt3+1}{3-1}$

$=\frac{2\sqrt6+3+\sqrt3+2\sqrt2+\sqrt3+1}{2}$

$=\sqrt6+3+\sqrt2+\sqrt3+2$

$=\sqrt{2} +\sqrt{3}+\sqrt{4}+\sqrt{6}$

= RHS

Hence, LHS=RHS.