`(sinalpha+secalpha)^(2)+(cosalpha+cosecalpha)^(2)=(K+secalphacosecalpha)^(2)`, then K = ?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation \((\sin \alpha + \sec \alpha)^2 + (\cos \alpha + \csc \alpha)^2 = (K + \sec \alpha \csc \alpha)^2\), we will use the substitution method by evaluating the expression for a specific angle. Let's take \(\alpha = 45^\circ\).

### Step 1: Substitute \(\alpha = 45^\circ\)
We know that:
- \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\)
- \(\sec 45^\circ = \csc 45^\circ = \sqrt{2}\)

Substituting these values into the left-hand side of the equation:

\[ \left(\sin 45^\circ + \sec 45^\circ\right)^2 + \left(\cos 45^\circ + \csc 45^\circ\right)^2 \]

This simplifies to:

\[ \left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)^2 + \left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)^2 \]

### Step 2: Simplify the expression
Calculating \(\frac{1}{\sqrt{2}} + \sqrt{2}\):

\[ \frac{1}{\sqrt{2}} + \sqrt{2} = \frac{1 + 2}{\sqrt{2}} = \frac{3}{\sqrt{2}} \]

Now we can substitute this back into the equation:

\[ \left(\frac{3}{\sqrt{2}}\right)^2 + \left(\frac{3}{\sqrt{2}}\right)^2 \]

### Step 3: Calculate the squares
Calculating the squares:

\[ \left(\frac{3}{\sqrt{2}}\right)^2 = \frac{9}{2} \]

Thus, we have:

\[ \frac{9}{2} + \frac{9}{2} = 9 \]

### Step 4: Evaluate the right-hand side
Now, we evaluate the right-hand side of the equation:

\[ (K + \sec 45^\circ \csc 45^\circ)^2 \]

Calculating \(\sec 45^\circ \csc 45^\circ\):

\[ \sec 45^\circ \csc 45^\circ = \sqrt{2} \cdot \sqrt{2} = 2 \]

So the right-hand side becomes:

\[ (K + 2)^2 \]

### Step 5: Set the two sides equal
Now we set the left-hand side equal to the right-hand side:

\[ 9 = (K + 2)^2 \]

### Step 6: Solve for \(K\)
Taking the square root of both sides gives us two equations:

\[ K + 2 = 3 \quad \text{or} \quad K + 2 = -3 \]

From \(K + 2 = 3\):

\[ K = 3 - 2 = 1 \]

From \(K + 2 = -3\):

\[ K = -3 - 2 = -5 \quad \text{(not valid since K is positive)} \]

Thus, the only valid solution is:

\[ K = 1 \]

### Final Answer
Therefore, the value of \(K\) is:

\[ \boxed{1} \]