If ` tantheta-cottheta=aandsintheta+costheta=b, then, (b^2-1)^2(a^2+4)` is equal to

Options

A. 2
B. `-4`
C. `pm 4`
D. 4

Explanation

To solve the problem, we need to find the value of \((b^2 - 1)^2 (a^2 + 4)\) given that \( \tan \theta - \cot \theta = a \) and \( \sin \theta + \cos \theta = b \).

### Step 1: Calculate \(b^2\)

We start with the expression for \(b\):
\[ b = \sin \theta + \cos \theta \]
Now, we square \(b\):
\[ b^2 = (\sin \theta + \cos \theta)^2 \]
Using the identity \((x + y)^2 = x^2 + 2xy + y^2\):
\[ b^2 = \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta \]
Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) and the double angle formula \(2\sin \theta \cos \theta = \sin 2\theta\):
\[ b^2 = 1 + \sin 2\theta \]

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

Now, we find \(b^2 - 1\):
\[ b^2 - 1 = \sin 2\theta \]

### Step 3: Calculate \((b^2 - 1)^2\)

Next, we square \(b^2 - 1\):
\[ (b^2 - 1)^2 = (\sin 2\theta)^2 = \sin^2 2\theta \]

### Step 4: Calculate \(a^2\)

Now we will calculate \(a^2\):
\[ a = \tan \theta - \cot \theta \]
We square \(a\):
\[ a^2 = (\tan \theta - \cot \theta)^2 \]
Using the identity \((x - y)^2 = x^2 - 2xy + y^2\):
\[ a^2 = \tan^2 \theta - 2\tan \theta \cot \theta + \cot^2 \theta \]
Since \(\tan \theta \cot \theta = 1\):
\[ a^2 = \tan^2 \theta - 2 + \cot^2 \theta \]
Using the identity \(\tan^2 \theta + \cot^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta}\):
\[ \tan^2 \theta + \cot^2 \theta = \frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta} \]
Thus:
\[ a^2 = \frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta} - 2 \]

### Step 5: Calculate \(a^2 + 4\)

Now, we find \(a^2 + 4\):
\[ a^2 + 4 = \frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta} - 2 + 4 = \frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta} + 2 \]

### Step 6: Calculate \((b^2 - 1)^2(a^2 + 4)\)

Now we multiply:
\[ (b^2 - 1)^2 (a^2 + 4) = \sin^2 2\theta \left( \frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta} + 2 \right) \]

Using the identity \(\sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta\):
\[ = \sin^2 2\theta \left( \frac{1 - 2\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} + 2 \right) \]
Since \(\sin^2 2\theta = 4\sin^2 \theta \cos^2 \theta\):
\[ = 4\sin^2 \theta \cos^2 \theta \left( \frac{1 - 2\sin^2 \theta \cos^2 \theta + 2\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \right) \]
This simplifies to:
\[ = 4 \]

### Final Answer

Thus, the value of \((b^2 - 1)^2(a^2 + 4)\) is:
\[ \boxed{4} \]