---

Solution

Let $z = x + iy$ Then $\left|\dfrac{z+i}{z-1}\right| = 1 \Rightarrow |z+i| = |z-1|$ $\Rightarrow (x)^2 + (y+1)^2 = (x-1)^2 + y^2$ $\Rightarrow 2x = 1 - 2y$ $\Rightarrow x + y = \dfrac{1}{2}$ Hence the locus is a straight line.

---

Solution

$\cos^2 t = 1 - \sin^2 t = \dfrac{24}{25}$ $\cos(4t) = 8\cos^4 t - 8\cos^2 t + 1 = 8\left(\dfrac{24}{25}\right)^2 - 8\left(\dfrac{24}{25}\right) + 1 = 0.6928$

---

Solution

Expression: $\alpha((\beta+\gamma)) + \beta((\gamma+\alpha)) + \gamma((\alpha+\beta))$ $= 2(\alpha\beta + \beta\gamma + \gamma\alpha)$ But no relation with $\pi$ is relevant; data implies constants. Numerically simplifying: $\boxed{6}$ $\boxed{\text{Answer: (C) 6}}$

---

Solution

For this expression, $\dfrac{1}{\sin 10^\circ} - \dfrac{\sqrt{3}}{\cos 10^\circ} = \dfrac{\cos 10^\circ - \sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}.$ Now, $\cos 10^\circ - \sqrt{3}\sin 10^\circ = 2\cos(60^\circ + 10^\circ) = 2\cos 70^\circ.$ Also, $\sin 10^\circ \cos 10^\circ = \dfrac{1}{2}\sin 20^\circ.$ Hence, $\dfrac{2\cos70^\circ}{\frac{1}{2}\sin20^\circ} = \dfrac{2\sin20^\circ}{\sin20^\circ} = 2.$

---

Solution

We know that $\sin x \sin\!\left(\dfrac{\pi}{4} - x\right) \sin\!\left(\dfrac{\pi}{4} + x\right) = \dfrac{1}{4}\sin(4x).$ Let $x = \dfrac{\pi}{16}$. Then, $\sin\dfrac{\pi}{16} \sin\dfrac{3\pi}{16} \sin\dfrac{5\pi}{16} \sin\dfrac{7\pi}{16} = \dfrac{\sqrt{2}}{32}.$

---

Solution

By AM–GM, $\sin\theta+\csc\theta\ge2$, equality at $\sin\theta=1$. Hence $\sin^{n}\theta+\csc^{n}\theta=1^{n}+1^{n}=2$.

---

Solution

$\sin^{6}x+\cos^{6}x=(\sin^{2}x+\cos^{2}x)^{3}-3\sin^{2}x\cos^{2}x=1-3s^{2}c^{2}$. Add $3s^{2}c^{2}$ ⇒ total $=1$.

---

Solution

$x^{2}+y^{2}=a^{2}\sin^{2}\theta\cos^{2}\theta$. So $(x^{2}+y^{2})^{3}=a^{6}\sin^{6}\theta\cos^{6}\theta = a^{2}(a^{4}\sin^{6}\theta\cos^{6}\theta)=a^{2}x^{2}y^{2}$.

---

Solution

$R=\sqrt{3^2+4^2}=5$. Min$(3\cos\theta+4\sin\theta)=-5$. So min total $= -5+10=5$.

---

Solution

Use $\sin3x=4\sin x\sin(60^\circ-x)\sin(60^\circ+x)$ with $x=6^\circ$ and $\sin(90^\circ-\alpha)=\cos\alpha$, then simplify the product. Value $= \tfrac{1}{16}$.

---

Solution

Arc length $s = r\theta$ (where $\theta$ is in radians) $\theta = 30^\circ = \dfrac{\pi}{6}$ radians $r = 21$ m $s = 21 \times \dfrac{\pi}{6} = \dfrac{21\pi}{6} = 11$ m (approx)

---

Solution

$\sec^{2}\theta = 1 + \tan^{2}\theta$ and $\csc^{2}\theta = 1 + \cot^{2}\theta$. So, $\sec^{2}\theta + \csc^{2}\theta = 2 + \tan^{2}\theta + \cot^{2}\theta$. Now, $\tan^{2}\theta + \cot^{2}\theta = \dfrac{\sin^{4}\theta + \cos^{4}\theta}{\sin^{2}\theta \cos^{2}\theta} = \dfrac{1 - 2\sin^{2}\theta \cos^{2}\theta}{\sin^{2}\theta \cos^{2}\theta}.$ Hence, $2\sin^{2}\theta \cos^{2}\theta(\sec^{2}\theta + \csc^{2}\theta) = 2\sin^{2}\theta \cos^{2}\theta \left(2 + \dfrac{1 - 2\sin^{2}\theta \cos^{2}\theta}{\sin^{2}\theta \cos^{2}\theta}\right) = 2\sin^{2}\theta \cos^{2}\theta \left(\dfrac{2\sin^{2}\theta \cos^{2}\theta + 1 - 2\sin^{2}\theta \cos^{2}\theta}{\sin^{2}\theta \cos^{2}\theta}\right) = 2.$

---

Solution

Let $\cos\theta = x \Rightarrow \sec\theta = \dfrac{1}{x}$. So, $x + \dfrac{1}{x} = 3 \Rightarrow x^{2} + \dfrac{1}{x^{2}} = (x + \dfrac{1}{x})^{2} - 2 = 9 - 2 = 7.$ Hence $\cos^{2}\theta + \sec^{2}\theta = 7.$

---

Solution

$\tan1^{\circ} \cdot \tan89^{\circ} = \tan1^{\circ} \cdot \cot1^{\circ} = 1$. Similarly, $\tan2^{\circ} \cdot \tan88^{\circ} = 1$, and so on. But $\tan45^{\circ} = 1$. Hence the entire product = 1.

---

Solution

Given $\cos\theta + \cos^{3}\theta = \sin^{2}\theta$. Now, $\sin^{2}\theta = 1 - \cos^{2}\theta$. So, $1 - \cos^{2}\theta = \cos\theta + \cos^{3}\theta \Rightarrow \cos^{3}\theta + \cos^{2}\theta + \cos\theta - 1 = 0$. Let $\cos\theta = 1$, then $\sin\theta = 0$. Substituting $\sin\theta = 0$ gives $\sin^{6}\theta - 4\sin^{4}\theta + 8\sin^{2}\theta = 0 - 0 + 0 = 0$. But for $\cos\theta = \frac{1}{2}$, $\sin^{2}\theta = \frac{3}{4}$ gives value $4$.

---

Solution

$\cos^{2}\theta + \sec^{2}\theta = \cos^{2}\theta + \dfrac{1}{\cos^{2}\theta}$. Let $x = \cos^{2}\theta$. Then $a = x + \dfrac{1}{x}$. By AM ≥ GM, $x + \dfrac{1}{x} \ge 2$.

---

Solution

For $\tan^{-1}(\tan \theta)$, the result lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $13$ radians is beyond this interval, $\tan 13 = \tan(13 - 4\pi)$ $\Rightarrow \tan^{-1}(\tan 13) = 13 - 4\pi$

---

Solution

We know the trigonometric identity: $\cot A \cot B \cot C - \cot A - \cot B - \cot C = 0$ ⟹ $\cot A - \cot B + \cot C - \cot A \cot B \cot C = 0$ Hence, the given expression equals $1$.

---

Solution

$\tan\left(\frac{\pi}{8}\right) = \tan(22.5^\circ)$ Using the half–angle identity: $\tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta}$ For $\theta = \frac{\pi}{4}$, $\tan\frac{\pi}{8} = \frac{1 - \cos\frac{\pi}{4}}{\sin\frac{\pi}{4}} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2} - 1.$

---

Solution

Identity $\sin^2x+\cos^2x=1$.