If `tantheta - cot theta =a` and `cos theta - sin theta =b`, then value of `(a^(2) + 4) (b^(2) -1)^(2)` is:

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we start with the given equations:

1. \( \tan \theta - \cot \theta = a \)
2. \( \cos \theta - \sin \theta = b \)

We need to find the value of \( (a^2 + 4)(b^2 - 1)^2 \).

### Step 1: Express \( a \) in terms of \( \theta \)

Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), we can rewrite \( a \):

\[ a = \tan \theta - \cot \theta = \frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\sin \theta} \]

Finding a common denominator:

\[ a = \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta} \]

### Step 2: Express \( b \) in terms of \( \theta \)

Now, we express \( b \):

\[ b = \cos \theta - \sin \theta \]

### Step 3: Substitute \( \theta = 45^\circ \)

To simplify our calculations, we can substitute \( \theta = 45^\circ \):

- At \( \theta = 45^\circ \):
- \( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \)

Calculating \( a \):

\[ a = \tan 45^\circ - \cot 45^\circ = 1 - 1 = 0 \]

Calculating \( b \):

\[ b = \cos 45^\circ - \sin 45^\circ = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0 \]

### Step 4: Calculate \( (a^2 + 4)(b^2 - 1)^2 \)

Now substituting \( a = 0 \) and \( b = 0 \):

\[ a^2 = 0^2 = 0 \]
\[ b^2 = 0^2 = 0 \]

Now substituting into the expression:

\[ (a^2 + 4)(b^2 - 1)^2 = (0 + 4)(0 - 1)^2 = 4 \cdot 1 = 4 \]

### Final Answer

Thus, the value of \( (a^2 + 4)(b^2 - 1)^2 \) is \( 4 \).

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