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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise Multiple Choice Questions Question 8 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise Multiple Choice Questions Question 8 Maths Textbook Solution.

Answers (1)

Answer: 2

Hint: We know the formula that, \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)

Given: \tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}

Solution:

          \begin{aligned} &\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ &\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4} \\ &\tan ^{-1}\left(\frac{2 x+3 x}{1-2 x(3 x)}\right)=\frac{\pi}{4} \\ &\frac{2 x+3 x}{1-2 x(3 x)}=\tan \frac{\pi}{4} \\ &\frac{5 x}{1-6 x^{2}}=1 \\ &5 x=1-6 x^{2} \\ &6 x^{2}+5 x-1=0 \end{aligned}

Therefore, there are 2 solutions as the formed equation is a quadratic equation.

\\ \\ \hspace{1cm}x=\frac{1}{6},\hspace{0.2cm}x=-1 (Rejected).

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