Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 1 maths Textbook Solution.

Answer: $x=\frac{-461}{9}$

Given: $\tan ^{-1}\left(\frac{1}{4}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$

Hint: Use formula

$\text { 1. } \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

$\text { 2. } 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$

Solution: We have,

$\tan ^{-1}\left(\frac{1}{4}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$

$\left[2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]$

$\Rightarrow \tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$

We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{\frac{1}{4}+\frac{5}{12}}{1-\frac{1}{4} \times \frac{5}{12}}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{8}{12}}{\frac{43}{48}}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{32}{43}\right)+\tan ^{-1}\left(\frac{1}{6}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4} \end{aligned}

$\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$

$\Rightarrow \tan ^{-1}\left(\frac{\frac{32}{43}+\frac{1}{6}}{1-\frac{32}{43} \times \frac{1}{6}}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{4}$                                    $\left[\tan ^{-1} \frac{\pi}{4}=1\right]$

$\Rightarrow \tan ^{-1}\left(\frac{\frac{235}{258}}{\frac{226}{258}}\right)+\tan ^{-1}\left(\frac{1}{\mathrm{x}}\right)=\tan ^{-1}(1)$

\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{235}{226}\right)+\tan ^{-1}\left(\frac{1}{x}\right)=\tan ^{-1}(1) \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{235}{226}+\frac{1}{x}}{1-\left(\frac{235}{226}\right) \times \frac{1}{x}}\right)=\tan ^{-1}(1), \frac{235}{226} x<1 \end{aligned}

\begin{aligned} &\Rightarrow \frac{235 x+226}{226 x-235}=1, x>\frac{235}{226} \\ &\Rightarrow 235 x+226=226 x-235, x>\frac{235}{226} \\ &\Rightarrow 235 x-226 x=-235-226 \\ &\Rightarrow 9 x=-461 \\ &\Rightarrow x=\frac{-461}{9} \end{aligned}

Hence $x=\frac{-461}{9}$is required answer