(i) If sin the `sec theta sec theta` =-1 and theta lies in the second quadrant find `"sin" theta` and `sec theta`
(ii) If `cos theta cosec theta` =- 1 and theta lies in the in the fourth quadrient find `cos theta and cosectheta`

Answer

Let's solve the two parts of the question step by step.

### Part (i)
**Given:**
\[ \sin \theta \sec \theta = -1 \]
**Condition:**
\(\theta\) lies in the second quadrant.

**Step 1:** Rewrite the equation using the identity for secant.
\[ \sec \theta = \frac{1}{\cos \theta} \]
Thus, the equation becomes:
\[ \sin \theta \cdot \frac{1}{\cos \theta} = -1 \]
This simplifies to:
\[ \frac{\sin \theta}{\cos \theta} = -1 \]
**Hint:** Remember that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).

**Step 2:** Recognize that \(\tan \theta = -1\).
Since \(\theta\) is in the second quadrant, where tangent is negative, we can conclude:
\[ \tan \theta = -1 \implies \theta = 135^\circ \quad (\text{or } 90^\circ + 45^\circ) \]

**Step 3:** Find \(\sin \theta\) and \(\sec \theta\).
Using the angle \(135^\circ\):
\[ \sin 135^\circ = \sin(90^\circ + 45^\circ) = \sin 45^\circ = \frac{1}{\sqrt{2}} \]
\[ \sec 135^\circ = \frac{1}{\cos 135^\circ} = \frac{1}{-\cos 45^\circ} = -\sqrt{2} \]

**Final Values:**
\[ \sin \theta = \frac{1}{\sqrt{2}}, \quad \sec \theta = -\sqrt{2} \]

### Part (ii)
**Given:**
\[ \cos \theta \csc \theta = -1 \]
**Condition:**
\(\theta\) lies in the fourth quadrant.

**Step 1:** Rewrite the equation using the identity for cosecant.
\[ \csc \theta = \frac{1}{\sin \theta} \]
Thus, the equation becomes:
\[ \cos \theta \cdot \frac{1}{\sin \theta} = -1 \]
This simplifies to:
\[ \frac{\cos \theta}{\sin \theta} = -1 \]
**Hint:** Remember that \(\cot \theta = \frac{\cos \theta}{\sin \theta}\).

**Step 2:** Recognize that \(\cot \theta = -1\).
Since \(\theta\) is in the fourth quadrant, where cotangent is negative, we can conclude:
\[ \cot \theta = -1 \implies \theta = 315^\circ \quad (\text{or } 360^\circ - 45^\circ) \]

**Step 3:** Find \(\cos \theta\) and \(\csc \theta\).
Using the angle \(315^\circ\):
\[ \cos 315^\circ = \cos(360^\circ - 45^\circ) = \cos 45^\circ = \frac{1}{\sqrt{2}} \]
\[ \sin 315^\circ = \sin(360^\circ - 45^\circ) = -\sin 45^\circ = -\frac{1}{\sqrt{2}} \]
Thus,
\[ \csc 315^\circ = \frac{1}{\sin 315^\circ} = \frac{1}{-\frac{1}{\sqrt{2}}} = -\sqrt{2} \]

**Final Values:**
\[ \cos \theta = \frac{1}{\sqrt{2}}, \quad \csc \theta = -\sqrt{2} \]

### Summary of Answers:
1. For part (i): \(\sin \theta = \frac{1}{\sqrt{2}}, \sec \theta = -\sqrt{2}\)
2. For part (ii): \(\cos \theta = \frac{1}{\sqrt{2}}, \csc \theta = -\sqrt{2}\)