A) \[tan\text{ }16\alpha \]
B) 0
C) \[\cot \alpha \]
D) none of these
Correct Answer: C
Solution :
\[tan\,\alpha +2\text{ }tan\text{ }2\alpha +4\text{ }tan\text{ }4\alpha +8\text{ }cot\text{ }8\alpha \] \[=tan\,\alpha +2\text{ }tan\text{ }2\alpha +4\text{ }tan\text{ }4\alpha +\frac{8}{\tan 8\alpha }\] \[=tan\text{ }\alpha +2tan\text{ }2\alpha +4\text{ }tan\text{ }4\alpha \] \[=tan\,\alpha +2tan\text{ }2\alpha \] \[+\left( \frac{4{{\tan }^{2}}4\alpha +4-4{{\tan }^{2}}4\alpha }{\tan 4\alpha } \right)\] \[=\tan \alpha +2\tan 2\alpha +\frac{4}{\tan 4\alpha }\] \[=\tan \alpha +2\tan 2\alpha +\frac{2(1-{{\tan }^{2}}2\alpha )}{\tan 2\alpha }\] \[=\tan \alpha +\left( \frac{2{{\tan }^{2}}2\alpha +2-2{{\tan }^{2}}2\alpha }{\tan 2\alpha } \right)\] \[=\tan \alpha +\frac{2}{\tan 2\alpha }=\tan \alpha +\frac{2(1-{{\tan }^{2}}\alpha )}{2\tan \alpha }\] \[=\frac{2{{\tan }^{2}}\alpha +2-2{{\tan }^{2}}\alpha }{2\tan \alpha }\] \[=\cot \alpha \] Alternative Method \[tan\text{ }\alpha +2tan\text{ }2\alpha +4\text{ }tan\text{ }4\alpha +8\text{ }cot\text{ }8\alpha \] \[=-[cot\,\alpha -tan\text{ }\alpha -2tan\text{ }2\alpha -4tan\text{ }4\alpha ]\] \[+8\text{ }cot\text{ }8\alpha +cot\text{ }\alpha \] \[=-[2cot\text{ }2\alpha -2tan\text{ }2\alpha -4\text{ }tan\text{ }4\alpha ]\] \[+8\,cot\,8\alpha +cot\text{ }\alpha \] \[=-[2(cot\text{ }2\alpha -tan\text{ }2\alpha )-4\text{ }tan\text{ }4\alpha ]\] \[\text{+}8\text{ }cot\text{ }8\alpha +cot\text{ }\alpha \] \[=-[2.2\text{ }cot\text{ }4\alpha -4\text{ }tan\text{ }4\alpha ]\] \[+8\text{ }cot\text{ }8\alpha +cot\text{ }\alpha \] \[=-4[cot\text{ }4\alpha -tan\text{ }4\alpha ]+8\text{ }cot\text{ }8\alpha +cot\text{ }\alpha \] \[=-\text{ }8\text{ }cot\text{ }8\alpha +8\text{ }cot\text{ }8\alpha +cot\alpha \] \[=cot\,\alpha \]
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