If `a=s e ctheta-t a ntheta` and `b=cos e ctheta+cottheta` , then show that `a b+a-b+1=0.`

Explanation

To show that \( ab + a - b + 1 = 0 \) given \( a = \sec \theta - \tan \theta \) and \( b = \csc \theta + \cot \theta \), we will follow these steps:

### Step 1: Write down the expression for \( ab + a - b + 1 \)

We start with the left-hand side (LHS):
\[ LHS = ab + a - b + 1 \]
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