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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Solve : \[ tan^{-1}(x+1)+tan^{-1}(x-1)=tan^{-1}\frac{8}{31}\]

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  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\bigg( \large\frac{x+y}{1-xy} \bigg) \)
By taking x+1 in the place of x and x-1 in the place of y we get
\(\frac{x+y}{1-xy}=\frac{(x+1)+(x-1)}{1-(x+1).(x-1)}=\frac{2x}{1-(x^2-1)}=\frac{2x}{2-x^2}\)
Substituting in the above formula of \(tan^{-1}x+tan^{-1}y\) we get
\(tan^{-1}(x+1)+tan^{-1}(x-1)=tan^{-1}\frac{2x}{2-x^2}=tan^{-1}\frac{8}{31}\)
\(\Rightarrow\:tan^{-1} \bigg(\large \frac{2x}{2-x^2} \bigg) = tan^{-1}\large\frac{8}{31}\)
\( \Rightarrow \large\frac{2x}{2-x^2}=\large\frac{8}{31}\)
\(\Rightarrow\:62x=8(2-x^2)\)
\(\Rightarrow\:31x=4(2-x^2)\)
\(\Rightarrow\:4x^2+31x-8=0\)
\(\Rightarrow\:4x^2+32x-x-8=0\)
\(\Rightarrow\:4x(x+8)-(x+8)=0\)
\(\Rightarrow\:(4x-1)(x+8)=0\)
Solving which we get\( x=-8\:\:or\:x=\large \frac{1}{4}\)
But \( x = -8\) is not in principal interval so \( x= \large\frac{1}{4}\)
answered Feb 28, 2013 by thanvigandhi_1
edited Mar 19, 2013 by rvidyagovindarajan_1
 
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