If \(\frac{1}{(\csc\theta+1)}+\frac{1}{(\csc\theta-1)}=2\sec\theta,\ 0^\circ<\theta<90^\circ\), then the value of \(\frac{\cot\theta-2\csc\theta}{\sec\theta}\) is:
\(\frac{\sqrt{2}-4}{2}\)
LHS = \(\frac{(\csc\theta-1)+(\csc\theta+1)}{\csc^2\theta-1} = \frac{2\csc\theta}{\cot^2\theta}\).
So \(\frac{2\csc\theta}{\cot^2\theta}=2\sec\theta\) ⇒ \(\frac{1}{\sin\theta}\cdot\frac{\sin^2\theta}{\cos^2\theta}=\frac{1}{\cos\theta}\) ⇒ \(\frac{\sin\theta}{\cos\theta}=1\).
tan θ = 1, so θ = 45°. Then cot θ = 1, cosec θ = \(\sqrt{2}\), sec θ = \(\sqrt{2}\).
Value = \(\frac{1-2\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}-4}{2}\) (multiplying numerator and denominator by \(\sqrt{2}\)).
Hence, the value is \(\frac{\sqrt{2}-4}{2}\).
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