Question

The Value of tan -1. 1/5 + tan -1. 1/3 + tan -1 1/3​

Answer 1

Answer:

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Answer 2

Solution -

$\: \: \: \sf{tan^{-1} \bigg(\dfrac{1}{5}\bigg) \: + \: tan^{-1} \bigg(\dfrac{1}{3}\bigg) \: + \: tan^{-1} \bigg(\dfrac{1}{3}\bigg)}$

$\: \: \: \sf{tan^{-1} \bigg(\dfrac{1}{5}\bigg) \: + \: 2 tan^{-1} \bigg(\dfrac{1}{3}\bigg)}$

as we know that -

$\: \: \: \: \: \: \sf{\boxed{\bold{\blue{2tan^{-1} x \: \: = \: \: tan^{-1} \dfrac{2x}{1 - x^{2}}}}}}$

On applying this property

↦$\: \: \sf{tan^{-1} \bigg(\dfrac{1}{5}\bigg) \: + \: tan^{-1} \bigg[\dfrac{2(\dfrac{1}{3})}{1 - (\dfrac{1}{3})^{2}}\bigg]}$

↦$\: \: \sf{tan^{-1} \bigg(\dfrac{1}{5}\bigg) \: + \: tan^{-1} \bigg[\dfrac{(\dfrac{2}{3})}{(1 - \dfrac{1}{9})}\bigg]}$

↦$\: \: \sf{tan^{-1} \bigg(\dfrac{1}{5}\bigg) \: + \: tan^{-1} \bigg[\dfrac{(\dfrac{\cancel{2}}{\cancel{3}})}{(\dfrac{4 \: \: \cancel{8}}{3 \: \: \cancel{9}})}\bigg]}$

↦$\: \: \sf{tan^{-1} \bigg(\dfrac{1}{5}\bigg) \: + \: tan^{-1} \bigg(\dfrac{3}{4}\bigg)}$

we also know that -

$\: \: \: \: \: \: \sf{\boxed{\bold{\blue{tan^{-1} x \: + \: tan^{-1} y \: \: = \: \: tan^{-1} \bigg(\dfrac{x + y}{1 - xy}\bigg)}}}}$

on applying this property

↦$\: \: \: \: \: \sf{tan^{-1} \bigg[\dfrac{(\dfrac{1}{5} \: + \: \dfrac{3}{4})}{1 \: - \: (\dfrac{1}{5})(\dfrac{3}{4})}\bigg]}$

↦$\: \: \: \: \: \: \sf{tan^{-1} \bigg[\dfrac{\dfrac{(4 + 15)}{20}}{(1 - \dfrac{3}{20})}\bigg]}$

↦$\: \: \: \: \: \: \: \: \: \sf{tan^{-1} \bigg[\dfrac{(\dfrac{19}{\cancel{20}})}{(\dfrac{17}{\cancel{20}})}\bigg]}$

↦$\: \: \: \: \: \: \: \: \: \sf{\boxed{\pink{tan^{-1} \bigg(\dfrac{19}{17}\bigg)}}}$