Indian Navy AA (Artificer Apprentice) Mathematics — practice by topic

Use the identity $\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\dfrac{a + b}{1 - ab}\right)$, valid when $ab < 1$. With $a = 1/2$, $b = 1/3$: $ab = 1/6 < 1$, so the identity applies cleanly. $\dfrac{a + b}{1 - ab} = \dfrac{1/2 + 1/3}{1 - 1/6} = \dfrac{5/6}{5/6} = 1$. So the sum equals $\tan^{-1}(1) = \dfrac{\pi}{4}$.

The **Pythagorean identity** $\sin^2\theta + \cos^2\theta = 1$ holds for every real $\theta$. It comes from Pythagoras applied to the unit circle: any point on it has coordinates $(\cos\theta, \sin\theta)$ with distance 1 from origin so the sum of squared coordinates is 1. Companion identities: - Dividing by $\cos^2\theta$: $1 + \tan^2\theta = \sec^2\theta$. - Dividing by $\sin^2\theta$: $\cot^2\theta + 1 = \csc^2\theta$.

For even $f$: $f(-x) = f(x)$, so the contributions from $[-a, 0]$ and $[0, a]$ are equal. **Thus $\int_{-a}^{a} f = 2\int_0^a f$.**