If cot α = \(\frac{1}{2}\),sec β = \(\frac{-5}{3}\),where π < α <

1.	If cot α = \(\frac{1}{2}\),sec β = \(\frac{-5}{3}\),where π < α < \(\frac{3\pi}{2}\) and \(\frac{\pi}{2}\) < β < π. Find value of tan (α + β).
Answer» Given, sec β = \(\frac{-5}{3}\) sec²β = \(\frac{25}{9}\) tan²β = \(\frac{25}{9}\)-1 = \(\frac{16}{9}\) tan β = ±\(\frac{4}{3}\) tan β = \(\frac{-4}{3}\) (\(\frac{\pi}{2}\)< β < π) And, cot α = \(\frac{1}{2}\) ∴ tan α = 2, (∵ \(\frac{\pi}{2}\)< α <\(\frac{3\pi}{2}\)) Now, tan (α + β) = \(\frac{tanα+tanβ}{1-tanα\,tanβ}\) \(= \frac{2+(\frac{-4}{3})}{1-2(\frac{-4}{3})}\) = \(\frac{6-4}{3+8}\) = \(\frac{2}{11}\)

If cot α = \(\frac{1}{2}\),sec β = \(\frac{-5}{3}\),where π < α < \(\frac{3\pi}{2}\) and \(\frac{\pi}{2}\) < β < π. Find value of tan (α + β).

Answer»

Given,

sec β = \(\frac{-5}{3}\)

sec²β = \(\frac{25}{9}\)

tan²β = \(\frac{25}{9}\)-1 = \(\frac{16}{9}\)

tan β = ±\(\frac{4}{3}\)

tan β = \(\frac{-4}{3}\) (\(\frac{\pi}{2}\)< β < π)

And, cot α = \(\frac{1}{2}\)

∴ tan α = 2, (∵ \(\frac{\pi}{2}\)< α <\(\frac{3\pi}{2}\))

Now,

tan (α + β) = \(\frac{tanα+tanβ}{1-tanα\,tanβ}\)

\(= \frac{2+(\frac{-4}{3})}{1-2(\frac{-4}{3})}\)

= \(\frac{6-4}{3+8}\)

= \(\frac{2}{11}\)