Question

Find the value of $ \tan \left( {{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) $

Answer 1

Hint: Convert $ {\cos ^{ - 1}} $ into $ {\tan ^{ - 1}} $ and then use the formula of inverse trigonometric functions to solve the question. We also use tan(A+B) to arrive at the result.

Complete step-by-step answer:
We need to find the value of
 $ \tan \left( {{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) $ . . . (1)
Let $ {\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) = \theta $
By taking cosec to both the sides, we get
 $ \cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right) = \cos \theta $
 $ \Rightarrow \dfrac{4}{5} = \cos \theta $ $ \left( {\because \cos ({{\cos }^{ - 1}}x) = x, - 1 \leqslant x \leqslant 1} \right) $
Rearranging it, we get
 $ \Rightarrow \cos \theta = \dfrac{4}{5} $
Taking square to both the sides, we get
 $ {\cos ^2}\theta = {\left( {\dfrac{4}{5}} \right)^2} $
 $ \Rightarrow {\cos ^2}\theta = \dfrac{{16}}{{25}} $
Taking reciprocal, we get
 $ \dfrac{1}{{{{\cos }^2}\theta }} = \dfrac{{25}}{{16}} $
 $ \Rightarrow {\sec ^2}\theta = \dfrac{{25}}{{16}} $ $ \left( {\because \sec \theta = \dfrac{1}{{\cos \theta }}} \right) $
 $ \Rightarrow 1 + {\tan ^2}\theta = \dfrac{{25}}{{16}} $ $ \left( {\because {{\sec }^2}\theta = 1 + {{\tan }^2}\theta } \right) $
Simplifying it, we get
 $ {\tan ^2}\theta = \dfrac{{25}}{{16}} - 1 $
 $ \Rightarrow {\tan ^2}\theta = \dfrac{{25 - 16}}{{16}} $
 $ \Rightarrow {\tan ^2}\theta = \dfrac{9}{{16}} $
By taking square root to both the sides, we get
 $ \Rightarrow \tan \theta = \dfrac{3}{4} $
Taking $ {\tan ^{ - 1}} $ to both the sides, we get
 $ {\tan ^{ - 1}}\left( {\tan \theta } \right) = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) $
 $ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) $ $ \left( {\because {{\tan }^{ - 1}}(\tan \theta ) = \theta } \right) $
Therefore, $ {\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) = \theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) $
Put this value in equation (1)
 $ \Rightarrow \tan \left( {{{\cos }^{ - 1}}\left( {\dfrac{3}{4}} \right) + {{\tan }^{ - 1}}\dfrac{2}{3}} \right) $
 $ = \tan \left( {{{\tan }^{ - 1}}\left( {\dfrac{3}{4}} \right) + {{\tan }^{ - 1}}\dfrac{2}{3}} \right) $
 $ = \tan \left\{ {{{\tan }^{ - 1}}\left( {\dfrac{{\dfrac{3}{4} + \dfrac{2}{3}}}{{1 - \dfrac{3}{4} \times \dfrac{2}{3}}}} \right)} \right\} $ $ \left( {\because {{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)} \right) $
By simplifying the above equation, we get
 $ = \tan \left\{ {{{\tan }^{ - 1}}\left( {\dfrac{{\dfrac{{9 + 8}}{{12}}}}{{1 - \dfrac{1}{2}}}} \right)} \right\} $
 $ = \tan \left\{ {{{\tan }^{ - 1}}\left( {\dfrac{{\dfrac{{17}}{{12}}}}{{\dfrac{1}{2}}}} \right)} \right\} $
 $ = \tan \left\{ {{{\tan }^{ - 1}}\left( {\dfrac{{17}}{{12}} \times 2} \right)} \right\} $
 $ = \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{17}}{6}} \right)} \right] $
 $ = \dfrac{{17}}{6} $ $ \left( {\because \tan ({{\tan }^{ - 1}}x) = x} \right) $
Therefore, the value of $ \tan \left( {{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{2}{3}} \right)} \right) = \dfrac{{17}}{6} $

Note: To solve such types of questions, first think about the formula that could be used to solve the question. Like in this question, all the terms were in $ \tan $ except one term. So it was clear that the formula that has $ {\tan ^{ - 1}} $ in it should be used. And for that, we converted $ {\cos ^{ - 1}} $ into $ {\tan ^{ - 1}} $ .