The value of `cot theta.tan (90^@ -theta)-sec(90^@ - theta) cosec theta+ (sin^2 25^@+sin^2 65^@) +sqrt3(tan 5^@ tan15^@.tan30^@.tan75^@.tan85^@)`is

Options

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

Answer

To solve the given expression step by step, we will break it down into manageable parts and utilize trigonometric identities.

### Given Expression:
\[ E = \cot \theta \tan (90^\circ - \theta) - \sec(90^\circ - \theta) \csc \theta + (\sin^2 25^\circ + \sin^2 65^\circ) + \sqrt{3} (\tan 5^\circ \tan 15^\circ \tan 30^\circ \tan 75^\circ \tan 85^\circ) \]

### Step 1: Simplify \(\tan(90^\circ - \theta)\) and \(\sec(90^\circ - \theta)\)
Using the identities:
- \(\tan(90^\circ - \theta) = \cot \theta\)
- \(\sec(90^\circ - \theta) = \csc \theta\)

We can rewrite the expression:
\[ E = \cot \theta \cdot \cot \theta - \csc \theta \cdot \csc \theta + (\sin^2 25^\circ + \sin^2 65^\circ) + \sqrt{3} (\tan 5^\circ \tan 15^\circ \tan 30^\circ \tan 75^\circ \tan 85^\circ) \]
\[ E = \cot^2 \theta - \csc^2 \theta + (\sin^2 25^\circ + \sin^2 65^\circ) + \sqrt{3} (\tan 5^\circ \tan 15^\circ \tan 30^\circ \tan 75^\circ \tan 85^\circ) \]

### Step 2: Use the identity \(\cot^2 \theta - \csc^2 \theta = -1\)
We know that:
\[ \cot^2 \theta - \csc^2 \theta = -1 \]
Thus, we can substitute this into our expression:
\[ E = -1 + (\sin^2 25^\circ + \sin^2 65^\circ) + \sqrt{3} (\tan 5^\circ \tan 15^\circ \tan 30^\circ \tan 75^\circ \tan 85^\circ) \]

### Step 3: Simplify \(\sin^2 25^\circ + \sin^2 65^\circ\)
Using the identity \(\sin(90^\circ - x) = \cos x\):
\[ \sin^2 65^\circ = \cos^2 25^\circ \]
Thus:
\[ \sin^2 25^\circ + \sin^2 65^\circ = \sin^2 25^\circ + \cos^2 25^\circ = 1 \]

### Step 4: Substitute back into the expression
Now substituting back:
\[ E = -1 + 1 + \sqrt{3} (\tan 5^\circ \tan 15^\circ \tan 30^\circ \tan 75^\circ \tan 85^\circ) \]
\[ E = \sqrt{3} (\tan 5^\circ \tan 15^\circ \tan 30^\circ \tan 75^\circ \tan 85^\circ) \]

### Step 5: Simplify \(\tan 75^\circ\) and \(\tan 85^\circ\)
Using the identities:
- \(\tan 75^\circ = \cot 15^\circ\)
- \(\tan 85^\circ = \cot 5^\circ\)

Thus:
\[ E = \sqrt{3} (\tan 5^\circ \tan 15^\circ \tan 30^\circ \cot 15^\circ \cot 5^\circ) \]
\[ E = \sqrt{3} (\tan 30^\circ) \]

### Step 6: Evaluate \(\tan 30^\circ\)
We know:
\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Thus:
\[ E = \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1 \]

### Final Answer:
\[ E = 1 \]