#### Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 8  Maths Textbook Solution.

Answer: $2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$

Hint: First we will solve for $2 \tan ^{-1} \frac{3}{4}$

Given: $2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$

Explanation:

L.H.S:

$2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}$                                                    $\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{3}{2}}{1-\frac{9}{16}}\right)-\tan ^{-1} \frac{17}{31} \\ &=\tan ^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right)-\tan ^{-1} \frac{17}{31} \\ &=\tan ^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)-\tan ^{-1} \frac{17}{31} \\ &=\tan ^{-1}\left(\frac{24}{7}\right)-\tan ^{-1} \frac{17}{31} \end{aligned}

$=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7} \times \frac{17}{31}}\right)$                                            $\left[\begin{array}{l} \because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right), x y>-1 \\ \frac{24}{7} \times \frac{17}{31}>-1 \end{array}\right]$

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{744-119}{217}}{1+\frac{408}{217}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right) \end{aligned}

\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}                                                                    $\left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]$

Hence it is proved that

$2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$