1.Ans: $x \in \{-1, 1\}$. Reason: $-1 \le \frac{x^2+1}{2x} \le 1 \Rightarrow |x^2+1| \le |2x| \Rightarrow x^2 - 2|x| + 1 \le 0 \Rightarrow (|x|-1)^2 \le 0$. This implies $|x| = 1$.
2.Ans: $4\pi - 10$. Reason: $10$ radians is approx $3.18\pi$. We know $\cos(4\pi - 10) = \cos 10$. Since $4\pi - 10 \approx 12.56 - 10 = 2.56 \in [0, \pi]$, this is the principal value.
3.Ans: $3\pi - 10$. Reason: $\sin(3\pi - 10) = \sin 10$. Since $3\pi - 10 \approx 9.42 - 10 = -0.58 \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, it is the valid principal value.
4.Ans: $x \in \left[-\frac{1}{4}, \frac{1}{2}\right]$. Reason: Inside the root must be $\ge 0$, so $\sin^{-1}(2x) \ge -\frac{\pi}{6} \Rightarrow 2x \ge -\frac{1}{2} \Rightarrow x \ge -\frac{1}{4}$. Also, domain of $\sin^{-1}(2x)$ requires $-1 \le 2x \le 1 \Rightarrow x \le \frac{1}{2}$.
5.Ans: Range is $[0, \pi]$. Reason: $f(x) = \sin^{-1}x + (\sin^{-1}x + \cos^{-1}x) = \sin^{-1}x + \frac{\pi}{2}$. The range of $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, so adding $\frac{\pi}{2}$ gives $[0, \pi]$.
6.Ans: $\frac{\pi}{4} - \frac{1}{2}\cos^{-1}(x^2)$. Reason: Substitute $x^2 = \cos 2\theta$. The expression simplifies to $\tan^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4}-\theta\right)\right) = \frac{\pi}{4} - \theta$.
7.Ans: $\frac{1+xy}{y-x}$. Reason: For $x, y > 1$, $\sin^{-1}\frac{2x}{1+x^2} = \pi - 2\tan^{-1}x$ and $\cos^{-1}\frac{1-y^2}{1+y^2} = 2\tan^{-1}y$. The expression becomes $\tan\left(\frac{\pi}{2} - \tan^{-1}x + \tan^{-1}y\right) = \cot(\tan^{-1}y - \tan^{-1}x)$.
8.Proof: $\cot^{-1}\left(\frac{ab+1}{a-b}\right) = \tan^{-1}\left(\frac{a-b}{1+ab}\right) = \tan^{-1}a - \tan^{-1}b$. Summing this cyclically gives $(\tan^{-1}a - \tan^{-1}b) + (\tan^{-1}b - \tan^{-1}c) + (\tan^{-1}c - \tan^{-1}a) = 0$.
9.Ans: $\tan^{-1}(n+1) - \frac{\pi}{4}$. Reason: Write term as $\tan^{-1}\left(\frac{(r+1)-r}{1+r(r+1)}\right) = \tan^{-1}(r+1) - \tan^{-1}r$. It becomes a telescoping series evaluating to $\tan^{-1}(n+1) - \tan^{-1}1$.
10.Ans: $\frac{3\pi}{4}$. Reason: $\frac{2}{n^2} = \frac{(n+1)-(n-1)}{1+(n+1)(n-1)}$. General term is $\tan^{-1}(n+1) - \tan^{-1}(n-1)$. Telescoping leaves $(\frac{\pi}{2} + \frac{\pi}{2}) - (\frac{\pi}{4} + 0) = \frac{3\pi}{4}$.
11.Ans: $3$. Reason: Max value of $\sin^{-1}$ is $\frac{\pi}{2}$. Thus $\sin^{-1}a = \sin^{-1}b = \sin^{-1}c = \frac{\pi}{2}$, meaning $a=b=c=1$. Plugging into expression gives $\frac{1+1+1}{1} = 3$.
12.Proof: Let $A = \cos^{-1}x, B = \cos^{-1}y, C = \cos^{-1}z$. $A+B = \pi - C \Rightarrow \cos(A+B) = -\cos C \Rightarrow xy - \sqrt{1-x^2}\sqrt{1-y^2} = -z$. Rearranging and squaring gives the result.
13.Ans: $\frac{3}{4}$. Reason: Let $\theta = \cos^{-1}\left(\frac{1}{8}\right) \Rightarrow \cos\theta = \frac{1}{8}$. We need $\cos\frac{\theta}{2} = \sqrt{\frac{1+\cos\theta}{2}} = \sqrt{\frac{9/8}{2}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.
14.Ans: $x + \tan^{-1}\left(\frac{5}{12}\right)$. Reason: Let $\sin\alpha = \frac{5}{13}, \cos\alpha = \frac{12}{13}$. The expression is $\sin^{-1}(\sin\alpha\cos x + \cos\alpha\sin x) = \sin^{-1}(\sin(x+\alpha)) = x + \alpha$.
15.Ans: Max: $\frac{5\pi^2}{4}$, Min: $\frac{\pi^2}{8}$. Reason: $f(x) = t^2 + (\frac{\pi}{2}-t)^2 = 2t^2 - \pi t + \frac{\pi^2}{4}$ where $t = \sin^{-1}x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Min at $t=\frac{\pi}{4}$, Max at $t=-\frac{\pi}{2}$.
16.Ans: $x \in \left(\frac{1}{\sqrt{2}}, 1\right]$. Reason: $\sin^{-1}x > \frac{\pi}{2} - \sin^{-1}x \Rightarrow 2\sin^{-1}x > \frac{\pi}{2} \Rightarrow \sin^{-1}x > \frac{\pi}{4}$.
17.Ans: $x = -\frac{1}{12}$. Reason: Both angles must be negative, so $x<0$. We get $6\sqrt{3}x = -\cos(\sin^{-1}6x) = -\sqrt{1-36x^2}$. Squaring gives $108x^2 = 1 - 36x^2 \Rightarrow 144x^2 = 1 \Rightarrow x = \pm\frac{1}{12}$.
18.Ans: $x = 0, \pm\frac{1}{2}$. Reason: Apply $\cos$ to both sides. $\cos(\cos^{-1}x - \sin^{-1}x) = x\sqrt{1-x^2} + \sqrt{1-x^2}x = 2x\sqrt{1-x^2}$. So $2x\sqrt{1-x^2} = x\sqrt{3}$, solving which gives roots.
19.Ans: $2$ solutions ($x=0, -1$). Reason: Domain forces $x(x+1) \ge 0$ and $x^2+x+1 \le 1 \Rightarrow x(x+1) \le 0$. Thus $x(x+1) = 0$. Both roots satisfy the equation.
20.Ans: $\frac{\pi}{5}$. Reason: Using $\cot^{-1}t = \frac{\pi}{2} - \tan^{-1}t$, we get $\pi - (\tan^{-1}x + \tan^{-1}y) = \pi - \frac{4\pi}{5} = \frac{\pi}{5}$.
21.Ans: $x = ab$. Reason: Rewrite as $\cos^{-1}\left(\frac{a}{x}\right) + \cos^{-1}a = \cos^{-1}\left(\frac{b}{x}\right) + \cos^{-1}b$. By inspection, substituting $x=ab$ perfectly balances the equation.
22.Ans: $x = \frac{3}{\sqrt{10}}$. Reason: $\cos^{-1}x = \frac{\pi}{4} - \sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$. $x = \cos\left(\frac{\pi}{4} - \alpha\right) = \cos\frac{\pi}{4}\cos\alpha + \sin\frac{\pi}{4}\sin\alpha = \frac{1}{\sqrt{2}}\left(\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}}\right)$.
23.Proof: $y = \frac{\pi}{2} - 2\tan^{-1}\sqrt{\cos x}$. Taking sine: $\sin y = \cos(2\tan^{-1}\sqrt{\cos x})$. Let $\tan^{-1}\sqrt{\cos x} = \theta$, then $\sin y = \cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \frac{1-\cos x}{1+\cos x} = \tan^2\left(\frac{x}{2}\right)$.
24.Ans: $x = 0, \pm\frac{1}{2}$. Reason: $\tan^{-1}(x-1) + \tan^{-1}(x+1) = \tan^{-1}3x - \tan^{-1}x$. Use subtraction/addition formulas: $\frac{2x}{2-x^2} = \frac{2x}{1+3x^2}$, which solves to $x=0$ or $x^2=\frac{1}{4}$.
25.Proof: $2\tan^{-1}y = \tan^{-1}x + \tan^{-1}z \Rightarrow \frac{2y}{1-y^2} = \frac{x+z}{1-xz}$. Since $x+z = 2y$ (A.P.), we have $2y\left(\frac{1}{1-y^2} - \frac{1}{1-xz}\right) = 0$. So $y=0$ or $y^2=xz$. If numbers are in AP and GP, they are equal.
26.Ans: No real solution. Reason: Max value of LHS is $2\pi/2 + \pi = 2\pi$, but domain restrictions on principal branches force the sum of these forms to strictly max out at $\pi$ for real $x$.
27.Ans: $x = -\frac{101}{3}$. Reason: Group terms and apply $\tan^{-1}A + \tan^{-1}B$ recursively.
28.Ans: $x \in \left[-\frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}\right] \cup \left[\frac{\sqrt{5}-1}{2}, \frac{1+\sqrt{5}}{2}\right]$. Reason: $|x - \frac{1}{x}| \ge 1$. Solve quadratic inequality $x^2 - x - 1 \ge 0$ and $x^2 + x - 1 \ge 0$ considering $x \neq 0$.
29.Proof: Apply cosine formula: $\frac{x}{a}\frac{y}{b} - \sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}} = \cos\alpha$. Isolate the radical, square both sides, and rearrange terms to get the required identity.
30.Ans: $\sqrt{\frac{x^2+1}{x^2+2}}$. Reason: Let $\tan^{-1}x = \theta \Rightarrow \tan\theta=x$. $\cos\theta = \frac{1}{\sqrt{x^2+1}}$. Now find $\sin(\cot^{-1}(\frac{1}{\sqrt{x^2+1}}))$. Let $\cot\phi = \frac{1}{\sqrt{x^2+1}} \Rightarrow \sin\phi = \frac{\text{perp}}{\text{hyp}} = \frac{\sqrt{x^2+1}}{\sqrt{1 + x^2 + 1}}$.
