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  • Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 5 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 5  Maths Textbook Solution.

Answers (1)

Answer: \sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}

Hints: First we will convert 2 \tan ^{-1}\left(\frac{1}{3}\right)in \sin ^{-1}

Given: \sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}

Explanation:

L.H.S:\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)                                                \left[\because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right]

=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{2 \times \frac{1}{3}}{1+\left(\frac{1}{3}\right)^{2}}\right)                                            \left[\begin{array}{c} -1 \leq x \leq 1 \\ -1 \leq \frac{1}{3} \leq 1 \end{array}\right]

\begin{aligned} &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{\frac{2}{3}}{1+\frac{1}{9}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{\frac{2}{3}}{\frac{10}{9}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{2}{3} \times \frac{9}{10}\right) \\ &=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{3}{5}\right) \end{aligned}

                                                                                                            \left[\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left\{x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right\}\right]

                                                                                                            \text { [if } \left.x, y \geq 0 \text { and } x^{2}+y^{2} \leq 1\right]

                                                                                                             \left[\begin{array}{l} \frac{4}{5}, \frac{3}{5} \geq 0,\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2} \\ \frac{16}{25}+\frac{9}{25}=\frac{25}{25}=1 \leq 1 \end{array}\right]

\begin{aligned} &=\sin ^{-1}\left\{\frac{4}{5} \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\frac{3}{5} \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right\} \\ &=\sin ^{-1}\left(\frac{4}{5} \sqrt{1-\frac{9}{25}}+\frac{3}{5} \sqrt{1-\frac{16}{25}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5} \sqrt{\frac{16}{25}}+\frac{3}{5} \sqrt{\frac{9}{25}}\right) \\ &=\sin ^{-1}\left(\frac{4}{5} \times \frac{4}{5}+\frac{3}{5} \times \frac{3}{5}\right) \\ &=\sin ^{-1}\left(\frac{16}{25}+\frac{9}{25}\right) \\ &=\sin ^{-1}\left(\frac{25}{25}\right) \end{aligned}

                                                                                                             \left[\begin{array}{l} \because \sin \frac{\pi}{2}=1 \\ \sin ^{-1}(1)=\frac{\pi}{2} \end{array}\right]

\begin{aligned} &=\sin ^{-1}(1) \\ &=\frac{\pi}{2} \end{aligned}

Hence it is proved that\sin ^{-1}\left(\frac{4}{5}\right)+2 \tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}

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