A) 0
B) 1/2
C) -1
D) 1
Correct Answer: D
Solution :
\[\tan \,({{100}^{o}}+{{125}^{o}})=\frac{\tan \,{{100}^{o}}+\tan \,{{125}^{o}}}{1-\tan \,{{100}^{o}}\,\tan \,{{125}^{o}}}\] \[\therefore \]\[\tan \,{{225}^{o}}=\frac{\tan \,\,{{100}^{o}}+\tan \,\,{{125}^{o}}}{1-\tan \,{{100}^{o}}\,\tan \,{{125}^{o}}}\] i.e., \[1=\frac{\tan \,{{100}^{o}}+\tan \,\,{{125}^{o}}}{1-\tan \,\,{{100}^{o}}\,\tan \,\,{{125}^{o}}}\] i.e.,\[\tan {{100}^{o}}+\tan {{125}^{o}}+\tan {{100}^{o}}\tan {{125}^{o}}=1.\]
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