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

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

Answers (1)

Answer: \tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1} \frac{4}{5}

Hints: First we will solve for \tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9} then we will convert it into \cos ^{-1} \text { and } \sin ^{-1} respectively.

Given:

\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1} \frac{4}{5}

First we will solve for \tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}

\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)

\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\left(\frac{1}{4}\right)\left(\frac{2}{9}\right)}\right)

                                      \begin{aligned} &=\tan ^{-1}\left(\frac{\frac{9+8}{36}}{1-\frac{1}{18}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{17}{36}}{\frac{17}{18}}\right) \\ &=\tan ^{-1}\left(\frac{17}{36} \times \frac{18}{17}\right) \end{aligned}

\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\tan ^{-1} \frac{1}{2}

                                      =\frac{1}{2} \times 2 \tan ^{-1} \frac{1}{2}                                        (On multiplying and dividing by 2)

\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \times \cos ^{-1}\left(\frac{1-\left(\frac{1}{2}\right)^{2}}{1+\left(\frac{1}{2}\right)^{2}}\right) \quad\left(\begin{array}{l} \because 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\ 0 \leq x<\infty \end{array}\right)

                                    \begin{aligned} &=\frac{1}{2} \cos ^{-1}\left(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}\right) \\ &=\frac{1}{2} \cos ^{-1}\left(\frac{\frac{3}{4}}{\frac{5}{4}}\right) \end{aligned}

\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right) .......................(1)

Now,let \cos ^{-1} \frac{3}{5}=\theta                             ....................(2)

\begin{aligned} &\cos \theta=\frac{3}{5} \\ &\therefore \sin \theta=\sqrt{1-\cos ^{2} \theta} \end{aligned}                                                \left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)

\begin{aligned} \sin \theta &=\sqrt{1-\cos ^{2} \theta} \\ \sin \theta &=\sqrt{1-\left(\frac{3}{5}\right)^{2}} \\ &=\sqrt{1-\frac{9}{25}} \\ &=\sqrt{\frac{25-9}{25}} \\ &=\sqrt{\frac{16}{25}} \end{aligned}

\begin{aligned} &\sin \theta=\frac{4}{5} \\ &\theta=\sin ^{-1} \frac{4}{5} \end{aligned}

from equation (2)

\cos ^{-1} \frac{3}{5}=\sin ^{-1} \frac{4}{5}                  .........(3)

From equation (1) and (2) we get

\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}=\frac{1}{2} \sin ^{-1} \frac{4}{5}

 

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