A) 2
B) 1
C) 0
D) 3
Correct Answer: A
Solution :
By \[\tan \,(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\] we have \[\tan {{70}^{o}}=\tan \,({{50}^{o}}+{{20}^{o}})\] \[\tan \,({{50}^{o}}+{{20}^{o}})=\frac{\tan {{50}^{o}}+\tan {{20}^{o}}}{1-\tan {{50}^{o}}\tan {{20}^{o}}}\] \[\therefore \] \[\tan {{70}^{o}}=\frac{\tan {{50}^{o}}+\tan {{20}^{o}}}{1-\tan {{50}^{o}}\tan {{20}^{o}}}\] \[\Rightarrow \] \[\tan {{70}^{o}}-(\tan {{50}^{o}}\tan {{20}^{o}})\,\tan {{70}^{o}}\] \[=\tan {{50}^{o}}+\tan {{20}^{o}}\] \[\Rightarrow \] \[\tan {{70}^{o}}-(\tan {{70}^{o}}\tan {{20}^{o}})\tan {{50}^{o}}\] \[=\tan {{50}^{o}}+\tan {{20}^{o}}\] \[\Rightarrow \] \[\tan {{70}^{o}}-\cot {{20}^{o}}\tan {{20}^{o}}\tan {{50}^{o}}\] \[=\tan {{50}^{o}}+\tan {{20}^{o}}\] [Using \[(90-\theta )=\cot \,\theta \]] \[\Rightarrow \] \[\tan {{70}^{o}}-\tan {{50}^{o}}=\tan {{50}^{o}}+\tan {{20}^{o}}\] \[[\tan \,\theta \,.\,\,\cot \theta =1]\] \[\Rightarrow \] \[\tan {{70}^{o}}-\tan {{20}^{o}}=2\tan {{50}^{o}}\] \[\Rightarrow \] \[\frac{\tan {{70}^{o}}-\tan {{20}^{o}}}{\tan {{50}^{o}}}=2\]
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