A) both a and b
B) band not a
C) a and not b
D) neither a nor b
Correct Answer: D
Solution :
\[{{\tan }^{-1}}\left( \frac{a}{b} \right)+{{\tan }^{-1}}\left( \frac{a+b}{a-b} \right)\] \[={{\tan }^{-1}}\left( \frac{\frac{a}{b}+\frac{a+b}{a-b}}{1-\frac{a}{b}\left( \frac{a+b}{a-b} \right)} \right)\] \[={{\tan }^{-1}}\left( \frac{{{a}^{2}}-ab+ab+{{b}^{2}}}{ab-{{b}^{2}}-{{a}^{2}}-ab} \right)\] \[={{\tan }^{-1}}\left( -\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \right)={{\tan }^{-1}}(-1)\] \[\therefore \]The value is neither depends on a nor b.
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