If `cosec^(2)theta=(25)/(16)and theta` is acute then what is the value of `(sqrt(cottheta+tantheta))`?

Options

A. a)`(sqrt(3))/(2)`
B. b)`(5)/(2sqrt(3))`
C. c)`(12)/(sqrt(3))`
D. d)`(2sqrt(3))/(3)`

Answer

To solve the problem, we start with the given equation:

\[ \csc^2 \theta = \frac{25}{16} \]

### Step 1: Find \(\sin^2 \theta\)

Using the identity \(\csc^2 \theta = \frac{1}{\sin^2 \theta}\), we can find \(\sin^2 \theta\):

\[ \sin^2 \theta = \frac{1}{\csc^2 \theta} = \frac{1}{\frac{25}{16}} = \frac{16}{25} \]

### Step 2: Find \(\cos^2 \theta\)

Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\):

\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \]

### Step 3: Find \(\tan \theta\) and \(\cot \theta\)

We know that:

\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]

Calculating \(\sin \theta\) and \(\cos \theta\):

\[ \sin \theta = \sqrt{\sin^2 \theta} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

\[ \cos \theta = \sqrt{\cos^2 \theta} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

Now we can find \(\tan \theta\):

\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]

Next, we find \(\cot \theta\):

\[ \cot \theta = \frac{1}{\tan \theta} = \frac{3}{4} \]

### Step 4: Calculate \(\sqrt{\cot \theta + \tan \theta}\)

Now we can calculate \(\cot \theta + \tan \theta\):

\[ \cot \theta + \tan \theta = \frac{3}{4} + \frac{4}{3} \]

To add these fractions, we need a common denominator, which is 12:

\[ \cot \theta + \tan \theta = \frac{3 \cdot 3}{4 \cdot 3} + \frac{4 \cdot 4}{3 \cdot 4} = \frac{9}{12} + \frac{16}{12} = \frac{25}{12} \]

Now we take the square root:

\[ \sqrt{\cot \theta + \tan \theta} = \sqrt{\frac{25}{12}} = \frac{5}{\sqrt{12}} = \frac{5}{2\sqrt{3}} \]

### Final Answer

Thus, the value of \(\sqrt{\cot \theta + \tan \theta}\) is:

\[ \frac{5}{2\sqrt{3}} \]