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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 2 Sub Question 9 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 2 Sub Question 9 Maths Textbook Solution.

Answers (1)

Answer: 2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}

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

Given: 2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}

Explanation:

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

                    \begin{aligned} &=\tan ^{-1}\left(\frac{1}{1-\left(\frac{1}{4}\right)}\right) \\ &=\tan ^{-1}\left(\frac{1}{3}{4}\right) \end{aligned}

2 \tan ^{-1} \frac{1}{2}=\tan ^{-1}\left(\frac{4}{3}\right)    ...................(1)

L.H.S:

2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}

From equation (1)

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

=\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \times \frac{1}{7}}\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{4}{3} \times \frac{1}{7}=\frac{4}{21}<1 \end{array}\right]

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{28+3}{21}}{1-\frac{4}{21}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{31}{21}}{\frac{17}{21}}\right) \\ &=\tan ^{-1}\left(\frac{31}{21} \times \frac{21}{17}\right) \\ &=\tan ^{-1}\left(\frac{31}{17}\right) \end{aligned}

Hence it is proved that 2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}

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