Evaluate ` sin (50^(@) + theta) - cos (40^(@) - theta) + tan 1^(@) tan 15^(@) tan 20^(@) tan 70^(@) tan 65^(@) tan 89^(@) + sec (90^(@) - theta). cosec theta - tan (90^(@) - theta ) cot theta `

Options

A. 0
B. 1
C. 2
D. 3

Answer

To evaluate the expression

\[ \sin(50^\circ + \theta) - \cos(40^\circ - \theta) + \tan(1^\circ) \tan(15^\circ) \tan(20^\circ) \tan(70^\circ) \tan(65^\circ) \tan(89^\circ) + \sec(90^\circ - \theta) \csc(\theta) - \tan(90^\circ - \theta) \cot(\theta), \]

we will simplify it step by step.

### Step 1: Simplify \(\sin(50^\circ + \theta) - \cos(40^\circ - \theta)\)

Using the identity \(\sin(A + B) = \cos(90^\circ - (A + B))\):

\[ \sin(50^\circ + \theta) = \cos(90^\circ - (50^\circ + \theta)) = \cos(40^\circ - \theta). \]

Thus,

\[ \sin(50^\circ + \theta) - \cos(40^\circ - \theta) = 0. \]

**Hint:** Use the sine and cosine identities to simplify expressions.

### Step 2: Simplify \(\sec(90^\circ - \theta) \csc(\theta)\)

Using the identity \(\sec(90^\circ - \theta) = \csc(\theta)\):

\[ \sec(90^\circ - \theta) \csc(\theta) = \csc(\theta) \csc(\theta) = \csc^2(\theta). \]

**Hint:** Remember that \(\sec(90^\circ - \theta) = \csc(\theta)\).

### Step 3: Simplify \(-\tan(90^\circ - \theta) \cot(\theta)\)

Using the identity \(\tan(90^\circ - \theta) = \cot(\theta)\):

\[ -\tan(90^\circ - \theta) \cot(\theta) = -\cot(\theta) \cot(\theta) = -\cot^2(\theta). \]

**Hint:** Recall that \(\tan(90^\circ - \theta) = \cot(\theta)\).

### Step 4: Combine the results

Now we have:

\[ 0 + \tan(1^\circ) \tan(15^\circ) \tan(20^\circ) \tan(70^\circ) \tan(65^\circ) \tan(89^\circ) + \csc^2(\theta) - \cot^2(\theta). \]

### Step 5: Evaluate \(\tan(1^\circ) \tan(15^\circ) \tan(20^\circ) \tan(70^\circ) \tan(65^\circ) \tan(89^\circ)\)

Using the identity \(\tan(90^\circ - x) = \cot(x)\), we can pair terms:

- \(\tan(1^\circ) = \cot(89^\circ)\)
- \(\tan(15^\circ) = \cot(75^\circ)\)
- \(\tan(20^\circ) = \cot(70^\circ)\)

Thus, we can express the product as:

\[ \tan(1^\circ) \tan(89^\circ) \cdot \tan(15^\circ) \tan(75^\circ) \cdot \tan(20^\circ) \tan(70^\circ) = 1 \cdot 1 \cdot 1 = 1. \]

**Hint:** Use the cotangent and tangent identities to simplify products.

### Step 6: Final evaluation

Now we can substitute back:

\[ 1 + \csc^2(\theta) - \cot^2(\theta). \]

Using the identity \(\csc^2(\theta) = 1 + \cot^2(\theta)\):

\[ 1 + (1 + \cot^2(\theta)) - \cot^2(\theta) = 1 + 1 = 2. \]

### Final Answer

Thus, the value of the expression is

\[ \boxed{2}. \]