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Question

Show that $$2{\tan}^{-1}\dfrac{1}{2}+{\tan}^{-1}\dfrac{1}{7}={\tan}^{-1}\dfrac{31}{17}$$.


Solution

We know that:
$$2\tan ^{ -1 }{ x } =\tan ^{ -1 }{ \dfrac { 2x }{ 1-{ x }^{ 2 } }  } $$

$$2\tan ^{ -1 }{ \dfrac { 1 }{ 2 }  } =\tan ^{ -1 }{ \dfrac { 2\times \frac { 1 }{ 2 }  }{ 1-{ \left( \dfrac { 1 }{ 2 }  \right)  }^{ 2 } }  } =\tan ^{ -1 }{ \dfrac { 4 }{ 3 }  } $$

Now, using the formula: 
$$\tan^{-1}A+\tan^{-1}B = \tan^{-1}\cfrac{A+B}{1-AB}$$

$$\tan ^{ -1 }{ \dfrac { 4 }{ 3 }  } +\tan ^{ -1 }{ \dfrac { 1 }{ 7 }  } =\tan ^{ -1 }{ \dfrac { \dfrac { 4 }{ 3 } +\dfrac { 1 }{ 7 }  }{ 1-\dfrac { 4 }{ 3 } \times \dfrac { 1 }{ 7 }  }  } =\tan ^{ -1 }{ \dfrac { 31 }{ 17 }  } $$

Mathematics

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