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  • Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise Multiple Choice Questions Question 25 Maths Textbbok Solution.

Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise Multiple Choice Questions Question 25 Maths Textbbok Solution.

Answers (1)

Answer: 2

Hint \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}

Given: \tan ^{-1} \frac{x+1}{x-1}+\tan ^{-1} \frac{x-1}{x}=\tan ^{-1}(-7)

Solution:

\therefore \tan ^{-1} \frac{x+1}{x-1}+\tan ^{-1} \frac{x-1}{x}=\tan ^{-1}(-7)

\tan ^{-1}\left(\frac{\frac{x+1}{x-1}+\frac{x-1}{x}}{1-\left(\frac{x+1}{x-1}\right)\left(\frac{x-1}{x}\right)}\right)=\tan ^{-1}(-7)

\tan ^{-1}\left(\frac{\frac{x^{2}+x+x^{2}-2 x+1}{x(x-1)}}{\frac{x^{2}-x-x^{2}+1}{x(x-1)}}\right)=\tan ^{-1}(-7)

\tan ^{-1}\left(\frac{2 x^{2}-x+1}{-x+1}\right)=\tan ^{-1}(-7)

\frac{2 x^{2}-x+1}{-x+1}=(-7)

\begin{aligned} &2 x^{2}-x+1=7 x-7 \\ &2 x^{2}-8 x+8=0 \\ &x^{2}-4 x+4=0 \end{aligned}

\begin{aligned} &(\mathrm{x}-2)^{2}=0 \\ &\mathrm{x}=2 \end{aligned}

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