UPSC NDA I 2024 Mathematics Paper

To find the value of the expression $\tan^{-1}\left(\frac{a}{b}\right) - \tan^{-1}\left(\frac{a-b}{a+b}\right)$, we can use the properties of the inverse tangent (or arctangent) function. Specifically, we will use the formula related to the difference of two arctangents:$\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$For our problem, let's set $x = \frac{a}{b}$ and $y = \frac{a-b}{a+b}$. Substituting these values into the formula above, we get:$\tan^{-1}\left(\frac{a}{b}\right) - \tan^{-1}\left(\frac{a-b}{a+b}\right) = \tan^{-1}\left(\frac{\frac{a}{b} - \frac{ab}{abb}}{1 + \frac{a}{b} \cdot \frac{ab}{a+b}}\right)$To simplify this expression, we need to find a common denominator for the numerator and simplify the denominator as follows:The numerator:$\frac{a}{b} - \frac{a-b}{a+b}$Finding a common denominator:$\frac{a(a+b)-b(a-b)}{b(a+b)} = \frac{a^2+ab-ab+b^2}{b(a+b)} = \frac{a^2+b^2}{b(a+b)}$The denominator:$1 + \frac{a}{b} \cdot \frac{a-b}{a+b}$Simplifying the product:$1 + \frac{a(a-b)}{b(a+b)} = 1 + \frac{a^2-ab}{b(a+b)} = \frac{b(a+b)+a^2-ab}{b(a+b)} = \frac{a^2+b^2}{b(a+b)}$Thus, we have: $\tan^{-1}\left(\frac{\frac{a^2+b^2}{b(a+b)}}{\frac{a^2+b^2}{b(a+b)}}\right) = \tan^{-1}(1)$We know that:$\tan^{-1}(1) = \frac{\pi}{4}$So:$\tan^{-1}\left(\frac{a}{b}\right) - \tan^{-1}\left(\frac{a-b}{a+b}\right) = \frac{\pi}{4}$