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What is the value of: \[tan^{-1} \bigg( \frac {x}{y} \bigg) -tan^{-1} \frac {x-y}{x+y} \]

\[ (A) \quad\frac{\pi}{2} \qquad (B) \quad \frac {\pi}{3}\qquad (C) \quad \frac {\pi}{4} \qquad (D)\quad \frac{-3\pi}{4} \]
Can you answer this question?
 
 

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  • \( tan^{-1}\alpha-tan^{-1}\beta=tan^{-1}\large\frac{\alpha-\beta}{1+\alpha\beta}\:\:\alpha\beta<1\)
Ans:  (C)   \(\frac{\pi}{4}\)
Given $ tan^{-1}\large \frac {x}{y} -tan^{-1}\large \frac {x-y}{x+y}$
 
We know that \( tan^{-1}\alpha-tan^{-1}\beta=tan^{-1}\large\frac{\alpha-\beta}{1+\alpha\beta}\:\:\alpha\beta<1\)
 
Let's take \(\alpha=\large\frac{x}{y}\:and\:\beta=\large\frac{x-y}{x+y}\)
\(1+ \alpha \beta=1+ \large \frac{x}{y} \frac{x-y}{x+y}= \large\frac{y(x+y)+x(x-y)}{y(x+y)} = \large\frac{xy+y^2+x^2-xy}{y(x+y)} = \large\frac{x^2+y^2}{y(x+y)} \)
\(\alpha -\beta=\large \frac{x}{y}-\large \frac{x-y}{x+y}=\large\frac{x(x+y)+y(x-y)}{y(x+y)} =\large \frac{x^2+xy-xy+y^2}{y(x+y)} = \large\frac{x^2+y^2}{y(x+y)} \)
 
Since $\alpha - \beta = 1+\alpha\;\beta \rightarrow tan^{-1}\large\frac{\alpha-\beta}{1+\alpha\beta} = \tan^{-1}1 = \large\frac{\pi}{4}$

 

answered Feb 28, 2013 by thanvigandhi_1
edited Mar 15, 2013 by rvidyagovindarajan_1
 

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