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  • Please solve RD Sharma class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 29 maths textbook solution

Please solve RD Sharma class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 29 maths textbook solution

Answers (1)

Answer:\frac{\pi}{4}

Given:\tan ^{-1} \frac{a}{b}-\tan ^{-1}\left(\frac{a-b}{a+b}\right)

Hint: \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)

Solution:         

\begin{aligned} \tan ^{-1} \frac{a}{b}-\tan ^{-1}\left(\frac{a-b}{a+b}\right) &=\tan ^{-1}\left(\frac{\frac{a}{b}-\frac{(a-b)}{a+b}}{1+\frac{a}{b}\left(\frac{a-b}{a+b}\right)}\right) \\\\ &=\tan ^{-1}\left(\frac{\frac{a^{2}+a b-a b+b^{2}}{b(a+b)}}{\frac{a b+b^{2}-a b+a^{2}}{b(a+b)}}\right) \\\\ &=\tan ^{-1}(1) \\\\ &=\tan ^{-1}\left(\tan \frac{\pi}{4}\right) \quad \therefore\left(\tan \frac{\pi}{4}=1\right) \\\\ &=\frac{\pi}{4} \end{aligned}

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