A) \[-2\]
B) \[1\]
C) \[-1\]
D) \[2\]
Correct Answer: B
Solution :
\[\cot \,{{12}^{o}}.\cot \,{{102}^{o}}+\cot \,{{102}^{o}}.\,\cot {{66}^{o}}+\cot \,{{66}^{o}}.\cot {{12}^{o}}\] \[\cot {{12}^{o}}.\,\cot ({{90}^{o}}+{{12}^{o}})+cot({{90}^{o}}+{{12}^{o}}){{.66}^{o}}+\cot {{66}^{o}}.\cot {{12}^{o}}\]\[=-\cot {{12}^{o}}.\,\tan {{12}^{o}}-\tan {{12}^{o}}.\cot {{66}^{o}}+\cot {{66}^{o}}.\cot {{12}^{o}}\] \[=-1+\cot {{66}^{o}}\{cot{{12}^{o}}-\tan {{12}^{o}}\}\] \[=-1+\cot 66.\left\{ \frac{1-{{\tan }^{2}}{{12}^{o}}}{\tan {{12}^{o}}} \right\}\] \[=-1+2\cot \,{{66}^{o}}.\left\{ \frac{{{\cot }^{2}}\,{{12}^{o}}-1}{2\cot \,{{12}^{o}}} \right\}\] \[=-1+2\cot \,{{66}^{o}}\cot {{24}^{o}}\] \[=-1+2\cot {{66}^{o}}.\cot ({{90}^{o}}-{{66}^{o}})\] \[=-1+2\cot {{66}^{o}}.\,\tan {{66}^{o}}\] \[=2-1=1\]
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