A) 1
B) 2
C) 3
D) 0
Correct Answer: B
Solution :
\[\frac{\tan {{70}^{o}}-\tan {{20}^{o}}}{\tan {{50}^{o}}}\] = \[\frac{\frac{\sin {{70}^{o}}}{\cos {{70}^{o}}}-\frac{\sin {{20}^{o}}}{\cos {{20}^{o}}}}{\frac{\sin {{50}^{o}}}{\cos {{50}^{o}}}}\]= \[\frac{\frac{\sin {{70}^{o}}\cos {{20}^{o}}-\cos {{70}^{o}}\sin {{20}^{o}}}{\cos {{70}^{o}}\cos {{20}^{o}}}}{\frac{\sin {{50}^{o}}}{\cos {{50}^{o}}}}\] = \[\frac{2}{2}\times \frac{\sin ({{70}^{o}}-{{20}^{o}})\cos {{50}^{o}}}{\cos {{70}^{o}}\cos {{20}^{o}}\sin {{50}^{o}}}\]= \[\frac{2\sin {{50}^{o}}\cos {{50}^{o}}}{2\cos {{70}^{o}}\cos {{20}^{o}}\sin {{50}^{o}}}\] = \[\frac{2\cos {{50}^{o}}}{\cos {{90}^{o}}+\cos {{50}^{o}}}=\frac{2\cos {{50}^{o}}}{0+\cos {{50}^{o}}}\] = 2.
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