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

Answers (1)

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<x<1

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


\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]



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