If `cos(alpha+beta)+sin(alpha-beta)=0` and `tan beta ne1`, then find the value of `tan alpha`.

Explanation

To solve the problem, we start with the equation given:

\[ \cos(\alpha + \beta) + \sin(\alpha - \beta) = 0 \]

### Step 1: Apply Trigonometric Identities

We can use the trigonometric identities for \(\cos(\alpha + \beta)\) and \(\sin(\alpha - \beta)\):

\[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]
\[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \]

Substituting these into the equation gives:

\[ \cos \alpha \cos \beta - \sin \alpha \sin \beta + \sin \alpha \cos \beta - \cos \alpha \sin \beta = 0 \]

### Step 2: Rearranging the Equation

Now, we can rearrange the equation:

\[ \cos \alpha \cos \beta - \cos \alpha \sin \beta + \sin \alpha \cos \beta - \sin \alpha \sin \beta = 0 \]

Grouping the terms:

\[ \cos \alpha (\cos \beta - \sin \beta) + \sin \alpha (\cos \beta - \sin \beta) = 0 \]

### Step 3: Factor Out Common Terms

We can factor out \((\cos \beta - \sin \beta)\):

\[ (\cos \beta - \sin \beta)(\cos \alpha + \sin \alpha) = 0 \]

### Step 4: Analyze the Factors

Since \(\tan \beta \neq 1\), we cannot have \(\cos \beta - \sin \beta = 0\). Therefore, we must have:

\[ \cos \alpha + \sin \alpha = 0 \]

### Step 5: Solve for \(\tan \alpha\)

From \(\cos \alpha + \sin \alpha = 0\), we can rearrange this to:

\[ \cos \alpha = -\sin \alpha \]

Dividing both sides by \(\cos \alpha\) (assuming \(\cos \alpha \neq 0\)) gives:

\[ 1 = -\tan \alpha \]

Thus:

\[ \tan \alpha = -1 \]

### Final Answer

The value of \(\tan \alpha\) is:

\[ \boxed{-1} \]