The value of `3(sin^4t+cos^4t-1)/(sin^6t+cos^6t-1)` is equal to __________

Explanation

To find the value of the expression

\[ \frac{3(\sin^4 t + \cos^4 t - 1)}{\sin^6 t + \cos^6 t - 1}, \]

we will follow these steps:

### Step 1: Simplify \(\sin^4 t + \cos^4 t\)

Using the identity \(\sin^2 t + \cos^2 t = 1\), we can square both sides:

\[ (\sin^2 t + \cos^2 t)^2 = 1^2. \]

Expanding this gives:

\[ \sin^4 t + \cos^4 t + 2\sin^2 t \cos^2 t = 1. \]

Rearranging this, we find:

\[ \sin^4 t + \cos^4 t = 1 - 2\sin^2 t \cos^2 t. \]

### Step 2: Substitute into the expression

Now substitute \(\sin^4 t + \cos^4 t\) back into the original expression:

\[ \sin^4 t + \cos^4 t - 1 = (1 - 2\sin^2 t \cos^2 t) - 1 = -2\sin^2 t \cos^2 t. \]

### Step 3: Simplify \(\sin^6 t + \cos^6 t\)

Using the identity for the sum of cubes, we have:

\[ \sin^6 t + \cos^6 t = (\sin^2 t + \cos^2 t)(\sin^4 t - \sin^2 t \cos^2 t + \cos^4 t). \]

Since \(\sin^2 t + \cos^2 t = 1\), we can simplify this to:

\[ \sin^6 t + \cos^6 t = \sin^4 t + \cos^4 t - \sin^2 t \cos^2 t. \]

Substituting \(\sin^4 t + \cos^4 t\) from Step 1:

\[ \sin^6 t + \cos^6 t = (1 - 2\sin^2 t \cos^2 t) - \sin^2 t \cos^2 t = 1 - 3\sin^2 t \cos^2 t. \]

### Step 4: Substitute into the denominator

Now substitute \(\sin^6 t + \cos^6 t\) back into the denominator:

\[ \sin^6 t + \cos^6 t - 1 = (1 - 3\sin^2 t \cos^2 t) - 1 = -3\sin^2 t \cos^2 t. \]

### Step 5: Substitute everything back into the expression

Now we can substitute everything back into the original expression:

\[ \frac{3(-2\sin^2 t \cos^2 t)}{-3\sin^2 t \cos^2 t}. \]

### Step 6: Simplify the expression

The \(-3\sin^2 t \cos^2 t\) cancels out:

\[ \frac{3 \cdot -2}{-3} = 2. \]

Thus, the value of the expression is

\[ \boxed{2}. \]