A) 0
B) \[\frac{1}{2}\]
C) -1
D) 1
Correct Answer: D
Solution :
[d]: Clearly, \[\tan ({{100}^{o}}+{{125}^{o}})=\frac{\tan {{100}^{o}}+\tan {{125}^{o}}}{1-\tan {{100}^{o}}\tan {{125}^{o}}}\] \[\Rightarrow \]\[\tan {{225}^{o}}=\frac{\tan {{100}^{o}}+\tan {{125}^{o}}}{1-\tan {{100}^{o}}\tan {{125}^{o}}}\] \[\Rightarrow \]\[1=\frac{\tan {{100}^{o}}+\tan {{125}^{o}}}{1-\tan {{100}^{o}}\tan {{125}^{o}}}\] \[\tan 100{}^\circ +\tan 125{}^\circ +\tan 100{}^\circ \tan 125{}^\circ =1\]
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