If `cos(2alpha+10^(@)).cosec(alpha-40^(@))=1` find `alpha`.

Answer

To solve the equation \( \cos(2\alpha + 10^\circ) \cdot \csc(\alpha - 40^\circ) = 1 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos(2\alpha + 10^\circ) \cdot \csc(\alpha - 40^\circ) = 1 \]
We can rewrite \(\csc(\alpha - 40^\circ)\) as \(\frac{1}{\sin(\alpha - 40^\circ)}\):
\[ \cos(2\alpha + 10^\circ) \cdot \frac{1}{\sin(\alpha - 40^\circ)} = 1 \]

### Step 2: Isolate \(\cos(2\alpha + 10^\circ)\)
Multiplying both sides by \(\sin(\alpha - 40^\circ)\) gives:
\[ \cos(2\alpha + 10^\circ) = \sin(\alpha - 40^\circ) \]

### Step 3: Use the co-function identity
We know that \(\sin(90^\circ - x) = \cos(x)\). We can express \(\sin(\alpha - 40^\circ)\) as:
\[ \sin(\alpha - 40^\circ) = \sin(90^\circ - (130^\circ - \alpha)) = \cos(130^\circ - \alpha) \]
Thus, we have:
\[ \cos(2\alpha + 10^\circ) = \cos(130^\circ - \alpha) \]

### Step 4: Set the angles equal
Since the cosines are equal, we can set the angles equal to each other:
\[ 2\alpha + 10^\circ = 130^\circ - \alpha \]

### Step 5: Solve for \(\alpha\)
Rearranging the equation gives:
\[ 2\alpha + \alpha = 130^\circ - 10^\circ \]
\[ 3\alpha = 120^\circ \]
Dividing both sides by 3:
\[ \alpha = 40^\circ \]

### Final Answer
Thus, the value of \(\alpha\) is:
\[ \alpha = 40^\circ \]

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