NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2

1. Put \(x = \sin\theta\), \[\theta = \sin^{-1}x \in \left[-\tfrac{\pi}{6}, \tfrac{\pi}{6}\right]\]
2. \[\begin{aligned}3x - 4x^3 &= 3\sin\theta - 4\sin^3\theta \\ &= \sin 3\theta\end{aligned}\]
3. \[3\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]\], the principal branch, so \[\begin{aligned}\sin^{-1}(\sin 3\theta) &= 3\theta \\ &= 3\sin^{-1}x\end{aligned}\]

1. Put \(x = \cos\theta\), \[\theta \in \left[0, \tfrac{\pi}{3}\right]\]
2. \[\begin{aligned}4x^3 - 3x &= 4\cos^3\theta - 3\cos\theta \\ &= \cos 3\theta\end{aligned}\]
3. \(3\theta \in [0, \pi]\), the principal branch of \(\cos^{-1}\), so \[\begin{aligned}\cos^{-1}(\cos 3\theta) &= 3\theta \\ &= 3\cos^{-1}x\end{aligned}\]

1. Put \(x = \tan\theta\), \[\theta \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\], \(\theta \ne 0\); then \(\sqrt{1 + x^2} = \sec\theta\) (positive, since \(\cos\theta > 0\)).
2. \[\begin{aligned}\dfrac{\sec\theta - 1}{\tan\theta} &= \dfrac{1 - \cos\theta}{\sin\theta} \\ &= \dfrac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \\ &= \tan\dfrac{\theta}{2}\end{aligned}\]
3. \[\tfrac{\theta}{2} \in \left(-\tfrac{\pi}{4}, \tfrac{\pi}{4}\right)\], so the expression is \(\tfrac{\theta}{2}\).

1. \[\begin{aligned}\dfrac{1 - \cos x}{1 + \cos x} &= \dfrac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}} \\ &= \tan^2\dfrac{x}{2}\end{aligned}\]
2. For \(0 < x < \pi\), \(\tan\tfrac{x}{2} > 0\), so the square root is \(\tan\tfrac{x}{2}\), and \[\tfrac{x}{2} \in \left(0, \tfrac{\pi}{2}\right)\]

1. \[\cos x - \sin x = \sqrt2\cos\left(x + \tfrac{\pi}{4}\right)\] and \[\cos x + \sin x = \sqrt2\sin\left(x + \tfrac{\pi}{4}\right)\], which is positive because \(x + \tfrac{\pi}{4} \in (0, \pi)\). (Dividing by \(\cos x\) instead would fail at \(x = \tfrac{\pi}{2}\), which is inside the interval.)
2. So the fraction is \[\begin{aligned}\cot\left(x + \tfrac{\pi}{4}\right) &= \tan\left(\tfrac{\pi}{2} - x - \tfrac{\pi}{4}\right) \\ &= \tan\left(\tfrac{\pi}{4} - x\right)\end{aligned}\]
3. \[\tfrac{\pi}{4} - x \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\] on the given interval, so \(\tan^{-1}\) returns it.

1. Put \(x = a\sin\theta\), \[\theta \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\]: \(\sqrt{a^2 - x^2} = a\cos\theta\).
2. The expression is \(\tan^{-1}(\tan\theta) = \theta\).

1. Put \(x = a\tan\theta\), \[\theta \in \left(-\tfrac{\pi}{6}, \tfrac{\pi}{6}\right)\]
2. \[\begin{aligned}\dfrac{3a^3\tan\theta - a^3\tan^3\theta}{a^3 - 3a^3\tan^2\theta} &= \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} \\ &= \tan 3\theta\end{aligned}\]
3. \[3\theta \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\], so the expression is \(3\theta\).

1. \(\sin^{-1}\tfrac12 = \tfrac{\pi}{6}\), so \(2\cos\tfrac{\pi}{3} = 1\).
2. \(\tan^{-1}1 = \tfrac{\pi}{4}\).

1. Put \(x = \tan\alpha\), \[\alpha \in \left(-\tfrac{\pi}{4}, \tfrac{\pi}{4}\right)\]: \(\dfrac{2x}{1 + x^2} = \sin 2\alpha\) with \[2\alpha \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\], so \[\sin^{-1}\dfrac{2x}{1 + x^2} = 2\tan^{-1}x\]
2. Put \(y = \tan\beta\), \(\beta \in \left(0, \tfrac{\pi}{2}\right)\): \(\dfrac{1 - y^2}{1 + y^2} = \cos 2\beta\) with \(2\beta \in (0, \pi)\), so \[\cos^{-1}\dfrac{1 - y^2}{1 + y^2} = 2\tan^{-1}y\]
3. The expression is \[\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\]

1. \(\tfrac{2\pi}{3}\) is not in \[\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]\], so the answer is not \(\tfrac{2\pi}{3}\).
2. \[\begin{aligned}\sin\tfrac{2\pi}{3} &= \sin\left(\pi - \tfrac{2\pi}{3}\right) \\ &= \sin\tfrac{\pi}{3}\end{aligned}\], and \(\tfrac{\pi}{3}\) is in the branch.

1. \[\begin{aligned}\tan\tfrac{3\pi}{4} &= \tan\left(\tfrac{3\pi}{4} - \pi\right) \\ &= \tan\left(-\tfrac{\pi}{4}\right)\end{aligned}\] (tan has period \(\pi\)), and \[-\tfrac{\pi}{4} \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\]

1. \(\sin^{-1}\tfrac35 = \alpha\): \(\sin\alpha = \tfrac35\), \(\cos\alpha = \tfrac45\), \(\tan\alpha = \tfrac34\). \(\cot^{-1}\tfrac32 = \beta\): \(\tan\beta = \tfrac23\).
2. \[\begin{aligned}\tan(\alpha + \beta) &= \dfrac{\frac34 + \frac23}{1 - \frac34 \cdot \frac23} \\ &= \dfrac{\frac{17}{12}}{\frac12}\end{aligned}\]

1. \(\tfrac{7\pi}{6} > \pi\) is outside \([0, \pi]\).
2. \[\begin{aligned}\cos\tfrac{7\pi}{6} &= \cos\left(2\pi - \tfrac{7\pi}{6}\right) \\ &= \cos\tfrac{5\pi}{6}\end{aligned}\], and \(\tfrac{5\pi}{6} \in [0, \pi]\).

1. \[\sin^{-1}\left(-\tfrac12\right) = -\tfrac{\pi}{6}\]
2. \[\begin{aligned}\sin\left(\tfrac{\pi}{3} + \tfrac{\pi}{6}\right) &= \sin\tfrac{\pi}{2} \\ &= 1\end{aligned}\]

1. \(\tan^{-1}\sqrt3 = \tfrac{\pi}{3}\); \(\cot^{-1}(-x) = \pi - \cot^{-1}x\), so \[\begin{aligned}\cot^{-1}\left(-\sqrt3\right) &= \pi - \tfrac{\pi}{6} \\ &= \tfrac{5\pi}{6}\end{aligned}\]
2. \[\tfrac{\pi}{3} - \tfrac{5\pi}{6} = -\tfrac{\pi}{2}\]

Where marks slip: The principal branch of \(\cot^{-1}\) is \((0, \pi)\), so \(\cot^{-1}(-\sqrt3) = \tfrac{5\pi}{6}\), not \(-\tfrac{\pi}{6}\). (Some software uses a different branch for \(\cot^{-1}\) of a negative number; the NCERT branch is \((0, \pi)\).)