A) \[-1\]
B) \[2\]
C) \[\frac{\pi }{2}\]
D) \[1\]
Correct Answer: D
Solution :
Given\[\tan {{1}^{o}}\tan {{2}^{o}}\tan {{3}^{o}}....\tan {{89}^{o}}\] Now, \[\tan {{89}^{o}}=\]of\[\tan (90-1)=\cot {{1}^{o}}\] \[\tan {{88}^{o}}=\cot {{2}^{o}}\]and so on. \[\therefore \]\[\tan {{1}^{o}}\tan {{2}^{o}}.....\tan {{89}^{o}}\] \[=(\tan {{1}^{o}}\cot {{1}^{o}})(tan{{2}^{o}}\cot {{2}^{o}}).....tan{{45}^{o}}\] \[1\cdot 1=1\]
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