Question

If $A=tan{6}^{0}tan{42}^0$ and $B=cot{66}^{0}cot{78}^0$, then

• A. A=2B
• B. A=1/3 B
• C. A= B
• D. 3A=2B

Answer 1

Answer: (A)

A=2B

Given $A=\tan 6\tan 42$ and $B=\cot 66\cot 78$

$\frac{A}{B}=\frac{\tan 6\tan 42}{\cot 66\cot 78}$

$=\frac{\sin 6\sin 42\sin 66\sin 78}{\cos 6\cos 42\cos 66\cos 78}$

$=\frac{(\cos 60-\cos 72)(\cos 36-\cos 120)}{(\cos 72+\cos 60)(\cos 120+\cos 36)}$

$=\frac{(\frac{1}{2}-\frac{\sqrt{5}-1}{4})(\frac{\sqrt{5}+1}{4}+\frac{1}{2})}{(\frac{1}{2}+\frac{\sqrt{5}-1}{4})(\frac{\sqrt{5}+1}{4}-\frac{1}{2})}$

$=\frac{\frac{1}{2}\left(\frac{\sqrt{5}+1\sqrt{5}+1}{4}\right)-\left(\frac{5-1}{16}\right)+\frac{1}{4}}{\frac{1}{2}\left(\frac{\sqrt{5}+1-\sqrt{5}+1}{4}\right)+\left(\frac{5-1}{16}\right)-\frac{1}{4}}$

$=\frac{\frac{2}{4}-\frac{1}{4}}{\frac{1}{4}}$

$=\frac{\frac{1}{4}}{\frac{1}{4}}$

$\frac{A}{B}=1$

$\therefore A=B$