#### Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 1 sub question 1 Maths Textbbok Solution.

Answer: $\frac{-7}{17}$

Hints: First we will solve for $2 \tan ^{-1} \frac{1}{5}$ and we use the formula

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

Given: $\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$

Explanation:

$\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$      ..............(1)

Let us first solve for $2 \tan ^{-1} \frac{1}{5}$

$2 \tan ^{-1} \frac{1}{5}=\tan ^{-1}\left\{\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right\}$                                        $\left[\because-1<\frac{1}{5}<1\right]$

$=\tan ^{-1}\left\{\frac{\frac{2}{5}}{1-\left(\frac{1}{25}\right)}\right\}$

$=\tan ^{-1}\left\{\frac{\frac{2}{5}}{\left(\frac{24}{25}\right)}\right\}$

$=\tan ^{-1}\left\{\frac{2}{5} \times \frac{25}{24}\right\}$

$2 \tan ^{-1} \frac{1}{5}=\tan ^{-1}\left(\frac{5}{12}\right)$................(2)

Now, we know that $\tan \frac{\pi}{4}=1$

$\frac{\pi}{4}=\tan ^{-1}(1)$

Now from equation (1), (2) and (3)

We get $\tan \left\{2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right\}$

$=\tan \left\{\tan ^{-1}\left(\frac{5}{12}\right)-\tan ^{-1}(1)\right\}$

$=\tan \left\{\tan ^{-1}\left(\frac{\frac{5}{12}-1}{1+\frac{5}{12} \times 1}\right)\right\}$                                                            $\left[\begin{array}{l} \because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right) \\ \text { forxy }>-1 \end{array}\right]$

$=\tan \left\{\tan ^{-1}\left(\frac{\frac{-7}{12}}{\frac{17}{12}}\right)\right\}$

$=\tan \left\{\tan ^{-1}\left(\frac{-7}{12} \times \frac{12}{17}\right)\right\}$

$=\tan \left\{\tan ^{-1}\left(\frac{-7}{17}\right)\right\}$                                                                            $\left[\because \tan \left(\tan ^{-1} \theta\right)=\theta\right]$

$=\frac{-7}{17}$