NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise

1. \[\begin{aligned}\cos\tfrac{13\pi}{6} &= \cos\left(2\pi + \tfrac{\pi}{6}\right) \\ &= \cos\tfrac{\pi}{6}\end{aligned}\], and \(\tfrac{\pi}{6} \in [0, \pi]\).

1. \[\begin{aligned}\tan\tfrac{7\pi}{6} &= \tan\left(\tfrac{7\pi}{6} - \pi\right) \\ &= \tan\tfrac{\pi}{6}\end{aligned}\], and \[\tfrac{\pi}{6} \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\]

1. Let \(\alpha = \sin^{-1}\tfrac35\): \(\cos\alpha = \tfrac45\), \(\tan\alpha = \tfrac34\), and \(\alpha < \tfrac{\pi}{4}\) (since \(\tfrac35 < \tfrac{1}{\sqrt2}\)).
2. \[\begin{aligned}\tan 2\alpha &= \dfrac{2 \cdot \frac34}{1 - \frac{9}{16}} \\ &= \dfrac{\frac32}{\frac{7}{16}} \\ &= \dfrac{24}{7}\end{aligned}\], and \[2\alpha \in \left(0, \tfrac{\pi}{2}\right)\], so \(2\alpha = \tan^{-1}\tfrac{24}{7}\).

1. \[\sin^{-1}\tfrac{8}{17} = \tan^{-1}\tfrac{8}{15}\] (right triangle 8, 15, 17); \(\sin^{-1}\tfrac35 = \tan^{-1}\tfrac34\).
2. \[\begin{aligned}\tan(\text{sum}) &= \dfrac{\frac{8}{15} + \frac34}{1 - \frac{8}{15} \cdot \frac34} \\ &= \dfrac{\frac{77}{60}}{\frac{36}{60}} \\ &= \dfrac{77}{36}\end{aligned}\]
3. Both angles are acute with product of tangents \(\tfrac25 < 1\), so the sum lies in \(\left(0, \tfrac{\pi}{2}\right)\) and equals \(\tan^{-1}\tfrac{77}{36}\).

1. \(\alpha = \cos^{-1}\tfrac45\): \(\sin\alpha = \tfrac35\). \(\beta = \cos^{-1}\tfrac{12}{13}\): \(\sin\beta = \tfrac{5}{13}\).
2. \[\begin{aligned}\cos(\alpha + \beta) &= \tfrac45 \cdot \tfrac{12}{13} - \tfrac35 \cdot \tfrac{5}{13} \\ &= \tfrac{48 - 15}{65} \\ &= \tfrac{33}{65}\end{aligned}\]
3. \(\alpha + \beta \in (0, \pi)\), so \(\alpha + \beta = \cos^{-1}\tfrac{33}{65}\).

1. \(\alpha = \cos^{-1}\tfrac{12}{13}\): \(\sin\alpha = \tfrac{5}{13}\). \(\beta = \sin^{-1}\tfrac35\): \(\cos\beta = \tfrac45\).
2. \[\begin{aligned}\sin(\alpha + \beta) &= \tfrac{5}{13} \cdot \tfrac45 + \tfrac{12}{13} \cdot \tfrac35 \\ &= \tfrac{20 + 36}{65} \\ &= \tfrac{56}{65}\end{aligned}\]
3. \[\begin{aligned}\cos(\alpha + \beta) &= \tfrac{12}{13} \cdot \tfrac45 - \tfrac{5}{13} \cdot \tfrac35 \\ &= \tfrac{33}{65} > 0\end{aligned}\], so \[\alpha + \beta \in \left(0, \tfrac{\pi}{2}\right)\] and equals \(\sin^{-1}\tfrac{56}{65}\).

1. \[\sin^{-1}\tfrac{5}{13} = \tan^{-1}\tfrac{5}{12}\]; \(\cos^{-1}\tfrac35 = \tan^{-1}\tfrac43\).
2. \[\begin{aligned}\tan(\text{sum}) &= \dfrac{\frac{5}{12} + \frac43}{1 - \frac{5}{12} \cdot \frac43} \\ &= \dfrac{\frac{21}{12}}{\frac{16}{36}} \\ &= \dfrac{63}{16}\end{aligned}\]
3. The product \[\tfrac{5}{12} \cdot \tfrac43 = \tfrac59 < 1\], so the sum is acute and equals \(\tan^{-1}\tfrac{63}{16}\).

1. Put \(x = \tan^2\theta\), \[\theta = \tan^{-1}\sqrt x \in \left[0, \tfrac{\pi}{4}\right]\]
2. \[\dfrac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta\], with \[2\theta \in \left[0, \tfrac{\pi}{2}\right] \subset [0, \pi]\]
3. So \[\begin{aligned}\tfrac12\cos^{-1}(\cos 2\theta) &= \theta \\ &= \tan^{-1}\sqrt x\end{aligned}\]

1. \[1 \pm \sin x = \left(\cos\tfrac{x}{2} \pm \sin\tfrac{x}{2}\right)^2\], and for \(x \in \left(0, \tfrac{\pi}{4}\right)\) both \(\cos\tfrac{x}{2} \pm \sin\tfrac{x}{2}\) are positive.
2. Numerator \(= 2\cos\tfrac{x}{2}\), denominator \(= 2\sin\tfrac{x}{2}\): the fraction is \(\cot\tfrac{x}{2}\).
3. \[\tfrac{x}{2} \in \left(0, \tfrac{\pi}{8}\right) \subset (0, \pi)\], so \[\cot^{-1}\left(\cot\tfrac{x}{2}\right) = \tfrac{x}{2}\]

1. Put \(x = \cos 2\theta\), \[2\theta = \cos^{-1}x \in \left[0, \tfrac{3\pi}{4}\right]\], so \[\theta \in \left[0, \tfrac{3\pi}{8}\right]\]
2. \(\sqrt{1 + x} = \sqrt2\cos\theta\), \(\sqrt{1 - x} = \sqrt2\sin\theta\) (both non-negative here).
3. Fraction \[= \dfrac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} = \tan\left(\tfrac{\pi}{4} - \theta\right)\], and \[\tfrac{\pi}{4} - \theta \in \left[-\tfrac{\pi}{8}, \tfrac{\pi}{4}\right]\] lies in the branch.
4. So the left side is \[\tfrac{\pi}{4} - \theta = \tfrac{\pi}{4} - \tfrac12\cos^{-1}x\]

1. \(\csc x\) needs \(\sin x \ne 0\), so \(|\cos x| < 1\), and \[2\tan^{-1}t = \tan^{-1}\dfrac{2t}{1 - t^2}\] for \(|t| < 1\). The equation becomes \[\dfrac{2\cos x}{1 - \cos^2 x} = \dfrac{2}{\sin x}\]
2. \[\begin{aligned}&\dfrac{2\cos x}{\sin^2 x} = \dfrac{2}{\sin x} \\ \Rightarrow\ &\cos x = \sin x \\ \Rightarrow\ &\tan x = 1\end{aligned}\]
3. The value in \((0, \pi)\) is \(x = \tfrac{\pi}{4}\); check: \[\begin{aligned}2\tan^{-1}\tfrac{1}{\sqrt2} &= \tan^{-1}\dfrac{\sqrt2}{1/2} \\ &= \tan^{-1}2\sqrt2 \\ &= \tan^{-1}\left(2\csc\tfrac{\pi}{4}\right)\end{aligned}\] (In general \(x = n\pi + \tfrac{\pi}{4}\), \(n \in \mathbb Z\), and each of these satisfies the equation.)

1. \[\begin{aligned}\tan^{-1}\dfrac{1 - x}{1 + x} &= \tan^{-1}1 - \tan^{-1}x \\ &= \tfrac{\pi}{4} - \tan^{-1}x\end{aligned}\] (for \(x > -1\)).
2. \[\begin{aligned}&\tfrac{\pi}{4} - \tan^{-1}x = \tfrac12\tan^{-1}x \\ \Rightarrow\ &\tfrac32\tan^{-1}x = \tfrac{\pi}{4} \\ \Rightarrow\ &\tan^{-1}x = \tfrac{\pi}{6}\end{aligned}\]

1. \(\theta = \tan^{-1}x\): a right triangle with opposite \(x\), adjacent 1, hypotenuse \(\sqrt{1 + x^2}\) (for \(x < 0\) the sign carries through, since \(\cos\theta > 0\)).
2. \(\sin\theta = \dfrac{x}{\sqrt{1 + x^2}}\).

1. \[\sin^{-1}(1 - x) = \tfrac{\pi}{2} + 2\sin^{-1}x\] Take sine: \(1 - x = \cos(2\sin^{-1}x) = 1 - 2x^2\), so \(2x^2 - x = 0\), \(x = 0\) or \(\tfrac12\).
2. Check \(x = \tfrac12\): \[\sin^{-1}\tfrac12 - 2\sin^{-1}\tfrac12 = -\tfrac{\pi}{6} \ne \tfrac{\pi}{2}\] Rejected.
3. Check \(x = 0\): \(\sin^{-1}1 - 0 = \tfrac{\pi}{2}\). Works.

Where marks slip: Squaring or taking sine can create extra roots: substitute each candidate back into the original equation.