If cosec θ + cot θ = K, 0° < θ < 90°, then find \(\frac{(K -1)^2 - 2}{(K+1)^2 - 2K}\) ?
Answer & explanation
Correct answer: option 1
Let K = 3
so triplets = (3, 4, 5)

Here,
⇒ cosec θ + cot θ
⇒ \(\frac{5}{3}\) + \(\frac{4}{3}\)
⇒ \(\frac{9}{3}\) = 3
Hence,
Put K = 3 in \(\frac{(K\;-\;1)^2\;-\;2}{(K\;+\;1)^2 -\;2K}\)
⇒ \(\frac{(3\;-\;1)^2\;-\;2}{(3\;+\;1)^2\;-\;2(3)}\)
= \(\frac{4\;-\;2}{16\;-\;6}\) = \(\frac{2}{10}\) = \(\frac{1}{5}\)
Put value in option
cos θ - sin θ = \(\frac{4}{5}\) - \(\frac{3}{5}\) = \(\frac{1}{5}\)