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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 4 Sub Question 1 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 4 Sub Question 1  Maths Textbook Solution.

Answers (1)

Answer: \tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}

Hint: First we will convert \cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right) into \tan ^{-1}

Given: \tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}

Explanation:

L.H.S:

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

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

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

\begin{aligned} &=\tan ^{-1}\left[\frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{1-1}\right] \\ &=\tan ^{-1}\left[\frac{\frac{1-x^{2}}{2 x}+\frac{2 x}{1-x^{2}}}{0}\right] \end{aligned}                                                                              \left(\frac{a}{0}=\infty\right)

=\tan ^{-1}(\infty)                                                                                                \left(\begin{array}{l} \because \tan \frac{\pi}{2}=\infty \\ \tan ^{-1}(\infty)=\frac{\pi}{2} \end{array}\right)

=\frac{\pi}{2}

Hence it is Proved that \tan ^{-1}\left(\frac{1-x^{2}}{2 x}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{2}

 

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