A) 2
B) \[\sqrt{2}\]
C) 4
D) 1
E) \[\frac{a}{b}\]
Correct Answer: B
Solution :
\[\sec \left[ {{\tan }^{-1}}\left( \frac{b+a}{b-a} \right)-{{\tan }^{-1}}\left( \frac{a}{b} \right) \right]\] \[=\sec \left[ {{\tan }^{-1}}\left\{ \frac{\frac{b+a}{b-a}-\frac{a}{b}}{\left( \frac{b+a}{b-a} \right)\left( \frac{a}{b} \right)} \right\} \right]\] \[=\sec [{{\tan }^{-1}}\{1\}]\] \[=\sec \frac{\pi }{4}=\sqrt{2}\]
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