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  • Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 4 Maths Textbbok Solution.

Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 4 Maths Textbbok Solution.

Answers (1)

Answer: \tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}

Hints: First we will use the formula of  2 \tan ^{-1} x.

Given: \tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}

Explanation:

                                                                                                    \left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{1}{3}<1 \end{array}\right]

 L.H.S:\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}                                      

=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right)

\begin{aligned} &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{\frac{2}{3}}{1-\frac{1}{9}}\right) \\ &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right) \\ &=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right) \end{aligned}                                    

                                                                                                    \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{3}{28}<1 \end{array}\right]

=\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{3}{4}\right)

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{3}{4}}{1-\left(\frac{1}{7}\right)\left(\frac{3}{4}\right)}\right) \\ &=\tan ^{-1}\left(\frac{\frac{4+21}{28}}{\frac{28-3}{28}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{25}{28}}{\frac{25}{28}}\right) \\ &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}                                                        \left[\begin{array}{l} \because \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]

Hence it is proved that \tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}

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