IIT JEE 1983 Maths - MCQ Question 4

$\begin{aligned} A & = \tan{\left[ \cos^{-1}{\dfrac{4}{5}}+\tan^{-1}{\dfrac{2}{3}}\right]} \\ & = \tan{\left[ \tan^{-1}{\dfrac{3}{4}}+\tan^{-1}{\dfrac{2}{3}}\right]} \\ & = \tan{\left[ \tan^{-1}{\left(\dfrac{\dfrac{3}{4}+\dfrac{2}{3}}{1-\dfrac{6}{12}}\right)} \right]} \quad \small \color{#3D99F6}{\frac{3}{4}\times\frac{2}{3} < 1} \\ & = \tan{\left(\tan^{-1}{\dfrac{17}{6}}\right)} \quad \quad \small \tan{\left(\tan^{-1}{x} \right)} = x \text{ for } x \in R \\ A & = \dfrac{17}{6}\end{aligned}$

By the identity $\displaystyle \tan \cos^{-1} x = \frac{\sin \cos ^{-1} x }{x}= \frac{\sqrt{1-\cos^2 \cos ^{-1} x }}{x}=\frac{\sqrt{1-x^2}}{x}$

Substituding $\displaystyle \tan\cos^{-1}\frac{4}{5} = \frac{3}{4}$ we have

$\displaystyle \tan\left(\cos^{-1}\frac{4}{5} + \tan^{-1}\frac{2}{3}\right) = \frac{\tan\cos^{-1}\frac{4}{5} + \frac{2}{3}}{1-\frac{2}{3}\tan\cos^{-1}\frac{4}{5}} = \frac{17}{6}$