The value of `cos24^0/(2tan33^0sin^2(57^0))+sin162^0/(sin18^0-cos18^0tan9^0)+cos162^0` is equal to

Answer

To solve the given expression:

\[ \frac{\cos 24^\circ}{2 \tan 33^\circ \sin^2(57^\circ)} + \frac{\sin 162^\circ}{\sin 18^\circ - \cos 18^\circ \tan 9^\circ} + \cos 162^\circ \]

we will simplify each term step by step.

### Step 1: Simplify \(\tan 33^\circ\) and \(\sin^2(57^\circ)\)

We know that:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Thus,
\[ \tan 33^\circ = \frac{\sin 33^\circ}{\cos 33^\circ} \]
Also, using the identity \(\sin(90^\circ - \theta) = \cos \theta\):
\[ \sin(57^\circ) = \cos(33^\circ) \]
Therefore,
\[ \sin^2(57^\circ) = \cos^2(33^\circ) \]

Now substituting these into the first term:
\[ \frac{\cos 24^\circ}{2 \cdot \frac{\sin 33^\circ}{\cos 33^\circ} \cdot \cos^2(33^\circ)} = \frac{\cos 24^\circ \cdot \cos 33^\circ}{2 \sin 33^\circ} \]

### Step 2: Simplify \(\sin 162^\circ\) and the denominator

Using the identity \(\sin(180^\circ - \theta) = \sin \theta\):
\[ \sin 162^\circ = \sin(180^\circ - 18^\circ) = \sin 18^\circ \]

Now substituting this into the second term:
\[ \frac{\sin 18^\circ}{\sin 18^\circ - \cos 18^\circ \tan 9^\circ} \]

Using \(\tan 9^\circ = \frac{\sin 9^\circ}{\cos 9^\circ}\):
\[ \cos 18^\circ \tan 9^\circ = \frac{\cos 18^\circ \sin 9^\circ}{\cos 9^\circ} \]

Thus, the denominator becomes:
\[ \sin 18^\circ - \frac{\cos 18^\circ \sin 9^\circ}{\cos 9^\circ} = \frac{\sin 18^\circ \cos 9^\circ - \cos 18^\circ \sin 9^\circ}{\cos 9^\circ} = \frac{\sin(18^\circ - 9^\circ)}{\cos 9^\circ} = \frac{\sin 9^\circ}{\cos 9^\circ} \]

So, the second term simplifies to:
\[ \frac{\sin 18^\circ \cos 9^\circ}{\sin 9^\circ} \]

### Step 3: Simplifying \(\cos 162^\circ\)

Using the identity \(\cos(180^\circ - \theta) = -\cos \theta\):
\[ \cos 162^\circ = -\cos 18^\circ \]

### Step 4: Combine all terms

Now we can combine all the simplified terms:
\[ \frac{\cos 24^\circ \cos 33^\circ}{2 \sin 33^\circ} + \frac{\sin 18^\circ \cos 9^\circ}{\sin 9^\circ} - \cos 18^\circ \]

### Step 5: Final simplification

We can use the half-angle formula:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
Thus:
\[ \sin 18^\circ = 2 \sin 9^\circ \cos 9^\circ \]

Substituting this back into the expression gives:
\[ \frac{\cos 24^\circ \cos 33^\circ}{2 \sin 33^\circ} + 2 \cos 9^\circ - \cos 18^\circ \]

### Conclusion

After simplifying and combining all terms, we find that the value of the expression is:
\[ \boxed{2} \]