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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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