Question

Solve for x the following :(a) $${ tan }^{ -1 }\dfrac { x-1 }{ x-2 } +{ tan }^{ -1 }\dfrac { x+1 }{ x+2 } =\dfrac { \pi }{ 4 }$$(b) $${ tan }^{ -1 }\dfrac { x-1 }{ x+1 } +{ tan }^{ -1 }\dfrac { 2x-1 }{ 2x+1 } ={ tan }^{ -1 }\dfrac { 23 }{ 36 }$$

Solution

(a)$$=\tan^{-1}\left ( \dfrac{x-1}{x-2} \right )+\tan^{-1}\left ( \dfrac{x+1}{x+2} \right )$$$$=\tan^{-1}\left ( \dfrac{\dfrac{x-1}{x-2}+\dfrac{x+1}{x+2}} {1-\dfrac{x-1}{x-2}\times \dfrac{x+1}{x+2}} \right )$$$$=\tan^{-1}\left ( \dfrac{2x^2-4} {-3} \right )$$$$=\dfrac{\pi}{4}$$$$\left ( \dfrac{2x^2-4} {-3} \right )=\tan^{-1}\dfrac{\pi}{4}$$$$\left ( \dfrac{2x^2-4} {-3} \right )=1$$$$2x^2-4=-3$$$$x^2=\dfrac{1}{2}$$$$\Rightarrow x=\dfrac{1}{\sqrt{2}}$$(b)$$=\tan^{-1}\left ( \dfrac{x-1}{x+1} \right )+\tan^{-1}\left ( \dfrac{2x-1}{2x+1} \right )$$$$=\tan^{-1}\left ( \dfrac{\dfrac{x-1}{x+1}+\dfrac{2x-1}{2x+1}} {1-\dfrac{x-1}{x+1}\times \dfrac{2x-1}{2x+1}} \right )$$$$=\tan^{-1}\left ( \dfrac{4x^2-2} {6x} \right )$$$$=\tan^{-1}\dfrac{23}{36}$$$$\left ( \dfrac{4x^2-2} {6x} \right )=\dfrac{23}{36}$$$$6(4x^2-2)=23x$$$$24x^2-23x-12 =0$$Solving the quadratic equation, we get,$$x=\dfrac {4} {3}$$ and $$x=\dfrac {-3} {8}$$Mathematics

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