PA and PB are two tangents drawn from point P to circle of radius 5 . A line is drawn from point P which cuts at C and D such that PC=5 and PD=15 and ` angleAPB= theta `.
On the basis of above information answer the questions .
value of sin2theta + cos4theta +sin5theta+tan7theta +cos 8theta

Options

A. `(4sqrt(3)-1)/(2)`
B. `(4sqrt(3)-1)/(2sqrt(3))`
C. `(4sqrt(3)+1)/(2sqrt(3))`
D. `(4sqrt(3)+1)/(2)`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the Geometry
We have a circle with a radius of 5 units. From point P, two tangents PA and PB are drawn to the circle. A line from point P intersects the points C and D such that PC = 5 units and PD = 15 units.

### Step 2: Calculate CD
Since PD = 15 and PC = 5, we can find the length of CD:
\[ CD = PD - PC = 15 - 5 = 10 \text{ units} \]

### Step 3: Identify the Triangle
The points A and B are where the tangents touch the circle. The radius at points A and B is perpendicular to the tangents. Therefore, triangles PCA and PCB are right triangles.

### Step 4: Use the Pythagorean Theorem
In triangle PCA, we can apply the Pythagorean theorem:
\[ PA^2 + PC^2 = PO^2 \]
Where PO is the distance from point P to the center O of the circle. Since PA is a tangent to the circle, we know:
\[ PA = \sqrt{PO^2 - OA^2} \]
Given OA (the radius) = 5 units, we need to find PO.

### Step 5: Find PO
From the triangle, we know:
\[ PO = PC + OA = 5 + 5 = 10 \text{ units} \]

### Step 6: Find angle θ
Using the right triangle PCA:
\[ \sin(\theta/2) = \frac{OA}{PO} = \frac{5}{10} = \frac{1}{2} \]
Thus,
\[ \theta/2 = 30^\circ \implies \theta = 60^\circ \]

### Step 7: Calculate the Required Expression
Now we need to calculate:
\[ \sin(2\theta) + \cos(4\theta) + \sin(5\theta) + \tan(7\theta) + \cos(8\theta) \]
Substituting \( \theta = 60^\circ \):
- \( \sin(2\theta) = \sin(120^\circ) = \frac{\sqrt{3}}{2} \)
- \( \cos(4\theta) = \cos(240^\circ) = -\frac{1}{2} \)
- \( \sin(5\theta) = \sin(300^\circ) = -\frac{\sqrt{3}}{2} \)
- \( \tan(7\theta) = \tan(420^\circ) = \tan(60^\circ) = \sqrt{3} \)
- \( \cos(8\theta) = \cos(480^\circ) = \cos(120^\circ) = -\frac{1}{2} \)

### Step 8: Combine the Results
Now, we combine these values:
\[ \sin(2\theta) + \cos(4\theta) + \sin(5\theta) + \tan(7\theta) + \cos(8\theta) = \frac{\sqrt{3}}{2} - \frac{1}{2} - \frac{\sqrt{3}}{2} + \sqrt{3} - \frac{1}{2} \]
Simplifying:
\[ = \left(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} + \sqrt{3}\right) + \left(-\frac{1}{2} - \frac{1}{2}\right) = \sqrt{3} - 1 \]

### Final Answer
Thus, the value of the expression is:
\[ \sqrt{3} - 1 \]

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