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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 3 Subquestion (ix) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 3 Subquestion (ix) Maths Textbook Solution.

Answers (1)

Answer:
\Rightarrow x=\pm 3
Hint:
Here, we will use the formula
\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )
Given:
\tan^{-1}\left ( 2+x \right )+\tan^{-1}\left ( 2-x\right )=\tan^{-1}\frac{2}{3}
Solution:
\begin{aligned} &\tan ^{-1}(2+x)+\tan ^{-1}(2-x)=\tan ^{-1} \frac{2}{3} \\ &\Rightarrow \tan ^{-1}\left(\frac{2+x+2-x}{1-(2+x)(2-x)}\right)=\tan ^{-1} \frac{2}{3} \\ &\Rightarrow \tan ^{-1}\left(\frac{4}{1-4+x^{2}}\right)=\tan ^{-1} \frac{2}{3} \\ &\Rightarrow \tan ^{-1}\left(\frac{4}{x^{2}-3}\right)=\tan ^{-1} \frac{2}{3} \end{aligned}
\! \! \! \! \! \! \! \! \! \Rightarrow 4x^{2}-3=23\\ \Rightarrow x^{2}-3=6\\ \Rightarrow x^{2}=9\\ \Rightarrow x=\pm 3\\

 

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