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 ${\tan ^{ - 1}}$ .