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  • Need Solution for R D Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 6 Maths Textbook Solution.

Need Solution for R D  Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 6 Maths Textbook Solution.

Answers (1)

Answer: \pi

Hint: First we will convert \sin ^{-1} \frac{2 x}{1+x^{2}} into \tan ^{-1}

Given: 2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^{2}} for x \geq 1

Explanation:

2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^{2}}                                                    \left[\because 2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}\right]

\begin{aligned} &=2 \tan ^{-1} x+2 \tan ^{-1} x \\ &=4 \tan ^{-1} x \quad(x \geq 1) \end{aligned}

Let x=1

                                                                                                    \left[\begin{array}{l} \because \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]

\begin{aligned} &=4 \tan ^{-1}(1) \\ &=4 \times \frac{\pi}{4} \\ &=\pi \end{aligned}

Hence the constant value for x \geq 1 is \pi

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