If `(sinalpha+"cosec"alpha)^(2)+(cosalpha+secalpha)^(2)=k+tan^(2)alpha+cot^(2)alpha` then the value of k is

Options

A. 1
B. 7
C. 3
D. 5

Answer

To solve the equation

\[ (\sin \alpha + \csc \alpha)^2 + (\cos \alpha + \sec \alpha)^2 = k + \tan^2 \alpha + \cot^2 \alpha \]

we will simplify the left-hand side step by step.

### Step 1: Expand the squares
Using the identity \((a + b)^2 = a^2 + b^2 + 2ab\), we expand both terms:

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

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

### Step 2: Substitute \(\csc \alpha\) and \(\sec \alpha\)
We know that \(\csc \alpha = \frac{1}{\sin \alpha}\) and \(\sec \alpha = \frac{1}{\cos \alpha}\). Therefore, we can rewrite:

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

### Step 3: Combine the expressions
Now, substituting these into the expanded squares:

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

This simplifies to:

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

### Step 4: Use the Pythagorean identity
Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\):

\[ 1 + \frac{1}{\sin^2 \alpha} + \frac{1}{\cos^2 \alpha} + 4 = 5 + \frac{1}{\sin^2 \alpha} + \frac{1}{\cos^2 \alpha} \]

### Step 5: Rewrite \(\frac{1}{\sin^2 \alpha}\) and \(\frac{1}{\cos^2 \alpha}\)
We can express \(\frac{1}{\sin^2 \alpha}\) as \(\cot^2 \alpha\) and \(\frac{1}{\cos^2 \alpha}\) as \(\tan^2 \alpha\):

\[ 5 + \tan^2 \alpha + \cot^2 \alpha \]

### Step 6: Set equal to the right-hand side
Now we have:

\[ 5 + \tan^2 \alpha + \cot^2 \alpha = k + \tan^2 \alpha + \cot^2 \alpha \]

### Step 7: Compare both sides
By comparing both sides, we can see that:

\[ k = 5 \]

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

\[ \boxed{5} \]