Question

$\operatorname{Cot} 12^{\circ} Cot 102^{\circ}+\operatorname{Cot} 102^{\circ} \operatorname{Cot} 66^{\circ}+\operatorname{Cot} 66^{\circ} \operatorname{Cot} 12^{\theta}=\ldots \ldots \ldots \ldots $

• A. $-1$
• B. $2$
• C. $-2$
• D. $1$

Answer 1

Answer: (D)

$\cot 12^{\circ} \cdot \cot 102^{\circ}+\cot 102^{\circ} \cdot \cot 66^{\circ}
+\cot 66^{\circ} \cdot \cot 12^{\circ}$
$ = \cot 12^{\circ} \cdot \cot (90^{\circ}+12^{\circ})+ \cot (90^{\circ}+12^{\circ}) \cdot \cot 66^{\circ} +\cot 66^{\circ} \cdot \cot 12^{\circ}$
$=-\cot 12^{\circ} \cdot \tan 12^{\circ}-\tan 12^{\circ} \cdot \cot 66^{\circ} +\cot 66^{\circ} \cdot \cot 12^{\circ}$
$=-1+\cot 66^{\circ} \cdot\left\{\cot 12^{\circ}-\tan 12^{\circ}\right\}$
$=-1+\cot 66 \cdot\left\{\frac{1-\tan ^{2} 12^{\circ}}{\tan 12^{\circ}}\right\}$
$=-1+2 \cot 66^{\circ} \cdot\left\{\frac{\cot ^{2} 12^{\circ}-1}{2 \cot 12^{\circ}}\right\}$
$=-1+2 \cot 66^{\circ} \cot 24^{\circ}$
$=-1+2 \cot 66^{\circ} \cdot \cot \left(90^{\circ}-66^{\circ}\right)$
$=-1+2 \cot 66^{\circ} \cdot \tan 66^{\circ}$
$=2-1=1$