The value of $(1 + \tan \theta + \sec \theta)(1 + \cot \theta - \csc \theta)$ is:
(a) $2$
(b) $1$
(c) $0$
(d) $-1$
Step-by-Step Solution
Key Concept: Convert to $\sin, \cos$: $\left(1 + \dfrac{\sin}{\cos} + \dfrac{1}{\cos}\right)\left(1 + \dfrac{\cos}{\sin} - \dfrac{1}{\sin}\right) = \dfrac{(\sin+\cos+1)(\sin+\cos-1)}{\sin \cos} = \dfrac{(\sin+\cos)^2 - 1}{\sin \cos} = \dfrac{2\sin\cos}{\sin\cos} = 2$.
$\dfrac{(\sin\theta + \cos\theta)^2 - 1}{\sin\theta \cos\theta} = \dfrac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = 2$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Evaluating product $= 2$: 1.0 Mark
Correct Answer: $2$