Q100 / 180 · KCET 2010 · Mathematics
$\cos 12^{\circ} \cot 102^{\circ}+\cot 102^{\circ} \cot 66^{\circ}$ $+\cot 66^{\circ} \cot 12^{\circ}$ is
- A$-2$
- Correct answerB1
- C$-1$
- D2
Solution
$\cot 12^{\circ} \cdot \cot 102^{\circ}+\cot 102^{\circ} \cdot \cot 66^{\circ}$
$+\cot 66^{\circ} \cdot \cot 12^{\circ}$
$\cot 12^{\circ} \cdot \cot \left(90^{\circ}+12^{\circ}\right)+\cot \left(90^{\circ}+12^{\circ}\right) \cdot 66^{\circ}$
$+\cot 66^{\circ} \cdot \cot 12^{\circ}$
$=-\cot 12^{\circ} \cdot \tan 12^{\circ}-\tan 12^{\circ} \cdot \cot 66^{\circ}$
$+\cot 66^{\circ} \cdot \cot 12^{\circ}$
$=-1+\cot 66^{\circ} \cdot\left\{\cot 12^{\circ}-\tan 12^{\circ}\right\}$
$=-1+\cot 66 \cdot\left\{\frac{1-\tan ^{2} 12^{\circ}}{\tan 12^{\circ}}\right\}$
$=-1+2 \cot 66^{\circ} \cdot\left\{\frac{\cot ^{2} 12^{\circ}-1}{2 \cot 12^{\circ}}\right\}$
$=-1+2 \cot 66^{\circ} \cot 24^{\circ}$
$=-1+2 \cot 66^{\circ} \cdot \cot \left(90^{\circ}-66^{\circ}\right)$
$=-1+2 \cot 66^{\circ} \cdot \tan 66^{\circ}$
$=2-1=1$
Hence, the correct answer is option: 'B'.