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Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 5 maths Textbook Solution.

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Answer: x=\sqrt{3}

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

Hint: Using formula

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

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


             We have,

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

We know that,

           \begin{aligned} &\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \\ &\Rightarrow \cos ^{-1}\left[\frac{-\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \end{aligned}                                                    [\cos (-\theta)=\pi-\cos \theta]

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

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

               \begin{aligned} &\Rightarrow \pi-2 \tan ^{-1}(x)+\tan ^{-1}(x)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-\tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\pi-\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow x=\tan \frac{\pi}{3} \end{aligned}

               \Rightarrow x=\sqrt{3}                                                                        \left[\tan \left(\frac{x}{3}\right)=\sqrt{3}\right]

   Hence x=\sqrt{3} is required solution.

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