If A,B,C,D be the angles of a quadrilateral , what is the value of `(sin(A+B))/(sin (C+D)) +(cos(C+D))/(cos (A+B))`

Answer

To solve the problem, we need to find the value of the expression:

\[ \frac{\sin(A+B)}{\sin(C+D)} + \frac{\cos(C+D)}{\cos(A+B)} \]

Given that \(A, B, C, D\) are the angles of a quadrilateral, we know that:

\[ A + B + C + D = 360^\circ \]

From this, we can express \(A + B\) in terms of \(C + D\):

\[ A + B = 360^\circ - (C + D) \]

### Step 1: Substitute \(A + B\)

Now, we substitute \(A + B\) into the expression:

\[ \frac{\sin(360^\circ - (C+D))}{\sin(C+D)} + \frac{\cos(C+D)}{\cos(360^\circ - (C+D))} \]

### Step 2: Use Trigonometric Identities

Using the trigonometric identities:

- \(\sin(360^\circ - \theta) = -\sin(\theta)\)
- \(\cos(360^\circ - \theta) = \cos(\theta)\)

We can rewrite the expression:

\[ \frac{-\sin(C+D)}{\sin(C+D)} + \frac{\cos(C+D)}{\cos(C+D)} \]

### Step 3: Simplify the Expression

Now, simplifying the expression gives:

\[ -1 + 1 \]

### Step 4: Final Result

Thus, the value of the entire expression simplifies to:

\[ 0 \]

### Final Answer:

The value of \(\frac{\sin(A+B)}{\sin(C+D)} + \frac{\cos(C+D)}{\cos(A+B)}\) is \(0\).

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