The value of `(sin25^(@)cos65^(@)+cos25^(@)sin65^(@))/(tan^(2)70^(@)-cosec^(2)20^(@))`

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To solve the expression \((\sin 25^{\circ} \cos 65^{\circ} + \cos 25^{\circ} \sin 65^{\circ}) / (\tan^{2} 70^{\circ} - \csc^{2} 20^{\circ})\), we can follow these steps:

### Step 1: Simplify the numerator
The numerator is \(\sin 25^{\circ} \cos 65^{\circ} + \cos 25^{\circ} \sin 65^{\circ}\). This expression can be recognized as the sine addition formula:
\[ \sin(a + b) = \sin a \cos b + \cos a \sin b \]
Here, \(a = 25^{\circ}\) and \(b = 65^{\circ}\). Therefore, we can rewrite the numerator as:
\[ \sin(25^{\circ} + 65^{\circ}) = \sin(90^{\circ}) = 1 \]

### Step 2: Simplify the denominator
The denominator is \(\tan^{2} 70^{\circ} - \csc^{2} 20^{\circ}\). We can use the identities for tangent and cosecant:
\[ \tan^{2} \theta = \frac{\sin^{2} \theta}{\cos^{2} \theta} \]
\[ \csc^{2} \theta = \frac{1}{\sin^{2} \theta} \]
Now, we know that \(70^{\circ} + 20^{\circ} = 90^{\circ}\), which means:
\[ \csc^{2} 20^{\circ} = \sec^{2} 70^{\circ} \]
Thus, we can rewrite the denominator:
\[ \tan^{2} 70^{\circ} - \sec^{2} 70^{\circ} \]

Using the identity \(\sec^{2} \theta = 1 + \tan^{2} \theta\), we find:
\[ \tan^{2} 70^{\circ} - \sec^{2} 70^{\circ} = \tan^{2} 70^{\circ} - (1 + \tan^{2} 70^{\circ}) = -1 \]

### Step 3: Combine the results
Now we can substitute back into the original expression:
\[ \frac{1}{-1} = -1 \]

### Final Answer
The value of the expression is \(-1\).

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