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  • Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 2 maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 2 maths Textbook Solution.

Answers (1)

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

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

Hint: Using formula since,

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

Solution: We have,

            3 \sin ^{-1} \frac{2 x}{1+x^{2}}-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}

We know that  2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\sin ^{-1} \frac{2 x}{1+x^{2}}=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)

           \begin{aligned} &\Rightarrow 3\left(2 \tan ^{-1} x\right)-4\left(2 \tan ^{-1} x\right)+2\left(2 \tan ^{-1} x\right)=\frac{\pi}{3} \\ &\Rightarrow 6 \tan ^{-1} x-8 \tan ^{-1} x+4 \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow 2 \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{6} \end{aligned}

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

         \Rightarrow x=\frac{1}{\sqrt{3}}

  This is required solution.

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