Without using trigonometric tables, show that `(cos 70^(@))/( sin 20^(@)) + cos 49^(@) cosec41^(@) = 2 `

Answer

To show that

\[ \frac{\cos 70^\circ}{\sin 20^\circ} + \cos 49^\circ \cdot \csc 41^\circ = 2 \]

we can follow these steps:

### Step 1: Rewrite \(\cos 70^\circ\)
Using the co-function identity, we know that:

\[ \cos 70^\circ = \sin(90^\circ - 70^\circ) = \sin 20^\circ \]

### Step 2: Substitute \(\cos 70^\circ\) in the expression
Now we can substitute \(\cos 70^\circ\) in the original expression:

\[ \frac{\sin 20^\circ}{\sin 20^\circ} + \cos 49^\circ \cdot \csc 41^\circ \]

### Step 3: Simplify the first term
The first term simplifies to:

\[ 1 + \cos 49^\circ \cdot \csc 41^\circ \]

### Step 4: Rewrite \(\cos 49^\circ\)
Using the co-function identity again, we can express \(\cos 49^\circ\) as:

\[ \cos 49^\circ = \sin(90^\circ - 49^\circ) = \sin 41^\circ \]

### Step 5: Substitute \(\cos 49^\circ\) in the expression
Now substitute \(\cos 49^\circ\) into the expression:

\[ 1 + \sin 41^\circ \cdot \csc 41^\circ \]

### Step 6: Simplify the second term
The term \(\sin 41^\circ \cdot \csc 41^\circ\) simplifies to:

\[ 1 \]

### Step 7: Combine the results
Now we can combine the results:

\[ 1 + 1 = 2 \]

Thus, we have shown that

\[ \frac{\cos 70^\circ}{\sin 20^\circ} + \cos 49^\circ \cdot \csc 41^\circ = 2 \]