If `(sin alpha + cosec alpha)^(2)+(cos alpha+sec alpha)^(2)=k+tan^(2)alpha+cot^(2)alpha`, then k=________

Options

A. 9
B. 7
C. 5
D. 3

Answer

To solve the equation \((\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = k + \tan^2 \alpha + \cot^2 \alpha\), we will follow these steps:

### Step 1: Expand the left-hand side
We start by expanding the squares on the left-hand side using the identity \((a + b)^2 = a^2 + b^2 + 2ab\).

\[ (\sin \alpha + \csc \alpha)^2 = \sin^2 \alpha + \csc^2 \alpha + 2 \sin \alpha \csc \alpha \]

\[ (\cos \alpha + \sec \alpha)^2 = \cos^2 \alpha + \sec^2 \alpha + 2 \cos \alpha \sec \alpha \]

### Step 2: Substitute \(\csc \alpha\) and \(\sec \alpha\)
Recall that \(\csc \alpha = \frac{1}{\sin \alpha}\) and \(\sec \alpha = \frac{1}{\cos \alpha}\). Thus, we can rewrite the expansions:

\[ \csc^2 \alpha = \frac{1}{\sin^2 \alpha}, \quad \sec^2 \alpha = \frac{1}{\cos^2 \alpha} \]

Now substituting these into our expansions gives:

\[ \sin^2 \alpha + \frac{1}{\sin^2 \alpha} + 2 \cdot 1 + \cos^2 \alpha + \frac{1}{\cos^2 \alpha} + 2 \cdot 1 \]

### Step 3: Combine the terms
Combining the terms, we have:

\[ \sin^2 \alpha + \cos^2 \alpha + \frac{1}{\sin^2 \alpha} + \frac{1}{\cos^2 \alpha} + 4 \]

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

\[ 1 + \frac{1}{\sin^2 \alpha} + \frac{1}{\cos^2 \alpha} + 4 \]

### Step 4: Express \(\tan^2 \alpha\) and \(\cot^2 \alpha\)
Recall that:

\[ \tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha}, \quad \cot^2 \alpha = \frac{\cos^2 \alpha}{\sin^2 \alpha} \]

Thus, we can express \(\tan^2 \alpha + \cot^2 \alpha\) as:

\[ \tan^2 \alpha + \cot^2 \alpha = \frac{\sin^2 \alpha + \cos^2 \alpha}{\sin^2 \alpha \cos^2 \alpha} = \frac{1}{\sin^2 \alpha \cos^2 \alpha} \]

### Step 5: Substitute back into the equation
Now we substitute this back into our equation:

\[ 1 + \frac{1}{\sin^2 \alpha} + \frac{1}{\cos^2 \alpha} + 4 = k + \frac{1}{\sin^2 \alpha \cos^2 \alpha} \]

### Step 6: Simplify the equation
Rearranging gives:

\[ 5 + \frac{1}{\sin^2 \alpha} + \frac{1}{\cos^2 \alpha} = k + \frac{1}{\sin^2 \alpha \cos^2 \alpha} \]

### Step 7: Isolate \(k\)
Now we isolate \(k\):

\[ k = 5 + \frac{1}{\sin^2 \alpha} + \frac{1}{\cos^2 \alpha} - \frac{1}{\sin^2 \alpha \cos^2 \alpha} \]

### Step 8: Use the identity
Using the identity \(\frac{1}{\sin^2 \alpha} + \frac{1}{\cos^2 \alpha} = \tan^2 \alpha + \cot^2 \alpha + 2\):

\[ k = 5 + 2 = 7 \]

Thus, the value of \(k\) is:

\[ \boxed{7} \]