Math, asked by sachintiwari3926, 1 year ago

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

Answers

Answered by Prajakta15
3
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
Answered by pinquancaro
1

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.

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