The value of `(tan 27^@ + cot 63^@)/(tan 27^@ (sin25^@ + cos65^@))`

Options

A. `cosec 25^@`
B. `2 tan 27^@`
C. `sin 25^@`
D. `tan 65^@`

Answer

To solve the expression \((\tan 27^\circ + \cot 63^\circ) / (\tan 27^\circ (\sin 25^\circ + \cos 65^\circ))\), we can follow these steps:

### Step 1: Simplify \(\cot 63^\circ\)
We know that \(\cot \theta = \frac{1}{\tan \theta}\). Therefore, we can write:
\[ \cot 63^\circ = \frac{1}{\tan 63^\circ} \]
Also, we know that \(\tan(90^\circ - \theta) = \cot \theta\), so:
\[ \tan 63^\circ = \cot 27^\circ \]
Thus, we have:
\[ \cot 63^\circ = \tan 27^\circ \]

### Step 2: Substitute \(\cot 63^\circ\) in the expression
Now substituting \(\cot 63^\circ\) in the original expression:
\[ \tan 27^\circ + \cot 63^\circ = \tan 27^\circ + \tan 27^\circ = 2\tan 27^\circ \]

### Step 3: Simplify \(\sin 25^\circ + \cos 65^\circ\)
We know that \(\cos 65^\circ = \sin(90^\circ - 65^\circ) = \sin 25^\circ\). Therefore:
\[ \sin 25^\circ + \cos 65^\circ = \sin 25^\circ + \sin 25^\circ = 2\sin 25^\circ \]

### Step 4: Substitute back into the expression
Now substituting back into the expression, we get:
\[ \frac{2\tan 27^\circ}{\tan 27^\circ (2\sin 25^\circ)} \]

### Step 5: Cancel out common terms
We can cancel \(2\tan 27^\circ\) in the numerator and denominator:
\[ = \frac{1}{\sin 25^\circ} \]

### Step 6: Final simplification
Using the identity \(\sin(90^\circ - \theta) = \cos \theta\), we can write:
\[ \sin 25^\circ = \cos 65^\circ \]
Thus, we have:
\[ \frac{1}{\sin 25^\circ} = \frac{1}{\cos 65^\circ} \]

### Final Answer
The value of the expression is:
\[ \frac{1}{\sin 25^\circ} = \sec 25^\circ \]

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