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Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 3 maths Textbook Solution.

Answers (1)

Answer: x=\frac{1}{\sqrt{3}}

Given: \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{2 \pi}{3}, x>0

Hint: Use

        \text { 1. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \text { and }

       \text { 2. } \cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x

Solution: We have,

         \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{2 \pi}{3}

We know that,

                    \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x and convert \cot ^{-1}into \tan ^{-1}

                   \Rightarrow 2 \tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}                                \left[\cot \theta=\frac{1}{\tan \theta}\right]

                   \Rightarrow 2 \tan ^{-1} x+2 \tan ^{-1} x=\frac{2 \pi}{3}                                               \left[\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x\right]

                   \begin{aligned} &\Rightarrow 4 \tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{2 \pi}{12} \\ &\Rightarrow x=\tan \frac{\pi}{6} \end{aligned}                                                                   \left[\tan ^{-1} x=\frac{\pi}{6} \Rightarrow x=\tan \frac{\pi}{6}\right]

                   \Rightarrow x=\frac{1}{\sqrt{3}}                                                                                \left[\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]

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