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  • Please solve RD Sharma class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 5 maths textbook solution

Please solve RD Sharma class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 5 maths textbook solution

Answers (1)

Answer: -2 \tan ^{-1} x

Given:

x<0, \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)

Hint:

\frac{1-\tan ^{2} x}{1+\tan ^{2} x}=\cos 2 x

Solution:

Let, x=\tan y\\\\

\begin{aligned} \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) &=\cos ^{-1}\left(\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\right) \\ &=\cos ^{-1}(\cos 2 y) \\ &=2 y-(1) \end{aligned}

The value of x  is negative, so let x=-a  where a>0 .           

\begin{array}{l} -a=\tan y \\\\ y=\tan ^{-1}(-a) \end{array}

 

Now,

\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 y          Using (1)

\begin{array}{l} =2 \tan ^{-1}(-a) \\\\ =2 \tan ^{1} x \quad[\therefore x=-a] \end{array}

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