I`(sinx+"cosec"x)^(2)+(cosx+secx)^(2)=k+tan^(2)x+cot^(2)x`, then what is the value of k ?

Options

A. 8
B. 7
C. 4
D. 3

Explanation

To solve the equation \( (\sin x + \csc x)^2 + (\cos x + \sec x)^2 = k + \tan^2 x + \cot^2 x \) and find the value of \( k \), we can follow these steps:

### Step 1: Expand the left-hand side
We start by expanding both terms on the left-hand side of the equation.

\[ (\sin x + \csc x)^2 = \sin^2 x + 2\sin x \csc x + \csc^2 x \]
Since \( \csc x = \frac{1}{\sin x} \), we have:
\[ \sin^2 x + 2 + \csc^2 x \]
And since \( \csc^2 x = 1 + \cot^2 x \):
\[ \sin^2 x + 2 + 1 + \cot^2 x = \sin^2 x + 3 + \cot^2 x \]

Now for the second term:
\[ (\cos x + \sec x)^2 = \cos^2 x + 2\cos x \sec x + \sec^2 x \]
Since \( \sec x = \frac{1}{\cos x} \), we have:
\[ \cos^2 x + 2 + \sec^2 x \]
And since \( \sec^2 x = 1 + \tan^2 x \):
\[ \cos^2 x + 2 + 1 + \tan^2 x = \cos^2 x + 3 + \tan^2 x \]

### Step 2: Combine the results
Now we combine the results from both expansions:
\[ (\sin^2 x + 3 + \cot^2 x) + (\cos^2 x + 3 + \tan^2 x) \]
This simplifies to:
\[ \sin^2 x + \cos^2 x + 6 + \tan^2 x + \cot^2 x \]
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \):
\[ 1 + 6 + \tan^2 x + \cot^2 x = 7 + \tan^2 x + \cot^2 x \]

### Step 3: Set the equation equal to the right-hand side
Now we have:
\[ 7 + \tan^2 x + \cot^2 x = k + \tan^2 x + \cot^2 x \]

### Step 4: Solve for \( k \)
By comparing both sides, we can see that:
\[ k = 7 \]

### Conclusion
Thus, the value of \( k \) is \( 7 \).