If ` cot theta=(4)/(3)`, then find the value of `sin theta,cos theta` and `cos ectheta` in first quadrant.
If `sin theta+cos ec theta=2` then find the value of `sin^(8 )theta+cos ec^(8)theta`.

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript and derive the required values.

### Step 1: Find sin θ, cos θ, and cosec θ

Given:
\[ \cot \theta = \frac{4}{3} \]

From the definition of cotangent:
\[ \cot \theta = \frac{\text{adjacent}}{\text{opposite}} = \frac{AB}{AC} \]
Let \( AB = 4 \) and \( AC = 3 \).

Using the Pythagorean theorem to find the hypotenuse \( BC \):
\[ BC = \sqrt{AB^2 + AC^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Now we can find \( \sin \theta \) and \( \cos \theta \):
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AC}{BC} = \frac{3}{5} \]
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{BC} = \frac{4}{5} \]

Now, we can find \( \csc \theta \):
\[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \]

### Step 2: Verify the values
We have:
- \( \sin \theta = \frac{3}{5} \)
- \( \cos \theta = \frac{4}{5} \)
- \( \csc \theta = \frac{5}{3} \)

### Step 3: Find \( \sin \theta + \csc \theta \)

Given:
\[ \sin \theta + \csc \theta = 2 \]
Substituting the values:
\[ \frac{3}{5} + \frac{5}{3} = 2 \]

To verify:
Finding a common denominator (15):
\[ \frac{3 \times 3}{5 \times 3} + \frac{5 \times 5}{3 \times 5} = \frac{9}{15} + \frac{25}{15} = \frac{34}{15} \neq 2 \]
This indicates that the assumption \( \sin \theta + \csc \theta = 2 \) is not satisfied with the initial values.

### Step 4: Find \( \sin^8 \theta + \csc^8 \theta \)

Since \( \sin \theta + \csc \theta = 2 \) holds true, we can square both sides:
\[ (\sin \theta + \csc \theta)^2 = 2^2 \]
Expanding:
\[ \sin^2 \theta + \csc^2 \theta + 2 \sin \theta \csc \theta = 4 \]
Since \( \csc \theta = \frac{1}{\sin \theta} \):
\[ \sin^2 \theta + \frac{1}{\sin^2 \theta} + 2 = 4 \]
Thus:
\[ \sin^2 \theta + \frac{1}{\sin^2 \theta} = 2 \]

Let \( x = \sin^2 \theta \):
\[ x + \frac{1}{x} = 2 \]
Multiplying through by \( x \):
\[ x^2 - 2x + 1 = 0 \]
Factoring:
\[ (x - 1)^2 = 0 \Rightarrow x = 1 \Rightarrow \sin^2 \theta = 1 \Rightarrow \sin \theta = 1 \]
Thus:
\[ \csc \theta = 1 \]

### Step 5: Calculate \( \sin^8 \theta + \csc^8 \theta \)
Since \( \sin \theta = 1 \) and \( \csc \theta = 1 \):
\[ \sin^8 \theta + \csc^8 \theta = 1^8 + 1^8 = 1 + 1 = 2 \]

### Final Answer:
\[ \sin^8 \theta + \csc^8 \theta = 2 \]