We are given the expression: \[ (1 + \cos x)(1 + \cot^2 x)(1 - \cos x) \] We know that: \[ 1 + \cot^2 x = \csc^2 x \] So the expression becomes: \[ (1 + \cos x)(1 - \cos x)(\csc^2 x) \] Now, use the identity: \[ (1 + \cos x)(1 - \cos x) = 1 - \cos^2 x = \sin^2 x \] Substituting back: \[ \sin^2 x \cdot \csc^2 x = \sin^2 x \cdot \frac{1}{\sin^2 x} = 1 \] So far it seems the value is 1. But the answer given is $-1$, which means there might be a misinterpretation. Let's re-evaluate. But wait, if we assume that $\cot^2 x = \cos^2 x / \sin^2 x$, then: \[ 1 + \cot^2 x = 1 + \frac{\cos^2 x}{\sin^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x} = \frac{1}{\sin^2 x} = \csc^2 x \] So the logic holds. Thus: \[ (1 + \cos x)(1 - \cos x)(\csc^2 x) = (1 - \cos^2 x)(\csc^2 x) = \sin^2 x \cdot \csc^2 x = 1 \]
The correct option is (B): \(1\)
The value of \(\dfrac{\sqrt{3}\cosec 20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}\) is equal to
If $\cot x=\dfrac{5}{12}$ for some $x\in(\pi,\tfrac{3\pi}{2})$, then \[ \sin 7x\left(\cos \frac{13x}{2}+\sin \frac{13x}{2}\right) +\cos 7x\left(\cos \frac{13x}{2}-\sin \frac{13x}{2}\right) \] is equal to
If \[ \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}, \] where \( \alpha, \beta \in \mathbb{N} \), then the value of \( \alpha + \beta \) is ___________.