Question

prove that: 2 tan-¹1/5+ tan-¹1/8= tan-¹4/7

Answer 1

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Answer 2

Question :

Prove that

$\sf2 \tan {}^{ - 1} \frac{1}{5} + \tan {}^{ - 1} \frac{1}{8} = \tan {}^{ - 1} \frac{4}{7}$

Formula's used :

$\sf1)2 \tan {}^{ - 1} x = \tan {}^{ - 1} ( \dfrac{2x}{1 - x {}^{2} } ) \: where \: - 1 < x < 1$

$\sf2) \tan { }^{ - 1} x + \tan {}^{ - 1} y = \tan {}^{ - 1} ( \dfrac{x + y}{1 - xy}) \: where \: xy < 1$

Solution :

We have to Prove that :

$\sf2 \tan {}^{ - 1} \frac{1}{5} + \tan {}^{ - 1} \frac{1}{8} = \tan {}^{ - 1} \frac{4}{7}$

LHS

$= \sf2 \tan {}^{ - 1} \frac{1}{5} + \tan {}^{ - 1} \frac{1}{8}$

$\sf \: use \:2 tan {}^{ - 1} x \: formula$

$= \sf \tan {}^{ - 1}( \dfrac{ \frac{2}{5} }{1 - (\frac{1}{5}){}^{2} }) + \tan {}^{ - 1} \frac{1}{8}$

$= \tan {}^{ - 1} ( \dfrac{ \frac{2}{5} }{1 - \frac{1}{25} } ) + \tan {}^{ - 1} \frac{1}{8}$

$\sf = \tan {}^{ - 1} ( \dfrac{10}{24} ) + \tan {}^{ - 1} ( \dfrac{1}{8})$

$\sf \: now \: use \: formula \tan {}^{ - 1} x + \tan {}^{ - 1}y$

$\sf = \tan {}^{ - 1} ( \dfrac{ \frac{10}{24} + \frac{1}{8} }{1 - \frac{10}{24} \times \frac{1}{8}} )$

$\sf = \tan {}^{ - 1} ( \dfrac{ \frac{10 + 3}{24} }{ \frac{24 \times 8 - 10}{24 \times 8} } )$

$\sf = \tan {}^{ - 1} ( \frac{13 \times 24 \times 8}{24 \times 182} )$

$\sf = \tan {}^{ - 1}( \frac{4}{7})$

RHS

$\sf = \tan {}^{ - 1} ( \frac{4}{7} )$

LHS = RHS

Hence proved !