If `2sec^2 alpha- sec^4 alpha-2 cosec^2 alpha+cosec^4 alpha=15/4` then `tan alpha=`

Options

A. `1//sqrt2`
B. `1//2`
C. `1//22sqrt2`
D. `-1//sqrt2`

Explanation

To solve the equation \( 2\sec^2 \alpha - \sec^4 \alpha - 2\csc^2 \alpha + \csc^4 \alpha = \frac{15}{4} \), we will follow these steps:

### Step 1: Rearranging the equation
We start by rearranging the equation to group the terms with squares and fourth powers together:

\[ 2\sec^2 \alpha - \sec^4 \alpha - 2\csc^2 \alpha + \csc^4 \alpha = \frac{15}{4} \]

### Step 2: Factoring the equation
We can factor the equation by recognizing that \( \csc^4 \alpha - \sec^4 \alpha \) can be expressed as a difference of squares:

\[ \sec^4 \alpha - \csc^4 \alpha = (\sec^2 \alpha - \csc^2 \alpha)(\sec^2 \alpha + \csc^2 \alpha) \]

Thus, we rewrite the equation as:

\[ 2\sec^2 \alpha - \sec^4 \alpha - 2\csc^2 \alpha + \csc^4 \alpha = 2(\sec^2 \alpha - \csc^2 \alpha) + (\csc^2 \alpha - \sec^2 \alpha)(\sec^2 \alpha + \csc^2 \alpha) = \frac{15}{4} \]

### Step 3: Substituting identities
Using the identities \( \sec^2 \alpha = 1 + \tan^2 \alpha \) and \( \csc^2 \alpha = 1 + \cot^2 \alpha \), we can substitute these into the equation:

\[ 2(1 + \tan^2 \alpha) - (1 + \tan^2 \alpha)^2 - 2(1 + \cot^2 \alpha) + (1 + \cot^2 \alpha)^2 = \frac{15}{4} \]

### Step 4: Simplifying the equation
Now we simplify the expression:

\[ 2 + 2\tan^2 \alpha - (1 + 2\tan^2 \alpha + \tan^4 \alpha) - 2 - 2\cot^2 \alpha + (1 + 2\cot^2 \alpha + \cot^4 \alpha) = \frac{15}{4} \]

This simplifies to:

\[ -\tan^4 \alpha + \cot^4 \alpha = \frac{15}{4} \]

### Step 5: Converting cotangent to tangent
Since \( \cot \alpha = \frac{1}{\tan \alpha} \), we can express \( \cot^4 \alpha \) in terms of \( \tan \alpha \):

\[ -\tan^4 \alpha + \frac{1}{\tan^4 \alpha} = \frac{15}{4} \]

### Step 6: Multiplying through by \( \tan^4 \alpha \)
Let \( x = \tan^4 \alpha \). Then we have:

\[ -x^2 + 1 = \frac{15}{4} x \]

Rearranging gives:

\[ 4x^2 + 15x - 4 = 0 \]

### Step 7: Solving the quadratic equation
We can solve this quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 4, b = 15, c = -4 \):

\[ x = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 4 \cdot (-4)}}{2 \cdot 4} \]
\[ x = \frac{-15 \pm \sqrt{225 + 64}}{8} \]
\[ x = \frac{-15 \pm \sqrt{289}}{8} \]
\[ x = \frac{-15 \pm 17}{8} \]

### Step 8: Finding the roots
Calculating the roots:

1. \( x = \frac{2}{8} = \frac{1}{4} \)
2. \( x = \frac{-32}{8} = -4 \) (not valid since \( x \) must be non-negative)

Thus, we have:

\[ \tan^4 \alpha = \frac{1}{4} \]

### Step 9: Taking the square root
Taking the square root gives:

\[ \tan^2 \alpha = \frac{1}{2} \]

Finally, taking the square root again:

\[ \tan \alpha = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]

### Final Answer
Thus, the value of \( \tan \alpha \) is:

\[ \tan \alpha = \frac{1}{\sqrt{2}} \text{ or } -\frac{1}{\sqrt{2}} \]