Question

The value of $ \tan {{1}^{o}}\tan {{2}^{o}}\tan {{3}^{o}}....tan{{89}^{o}} $ is equal to

• A. $ -1 $
• B. $ 2 $
• C. $ \frac{\pi }{2} $
• D. $ 1 $

Answer 1

Answer: (D)

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 $