The general solution of tan 5θ = cot 2θ is(a) θ = \(\frac{nπ}{7}

1.	The general solution of tan 5θ = cot 2θ is(a) θ = \(\frac{nπ}{7}\) + \(\frac{π}{14}\)(b) θ = \(\frac{nπ}{7}\) + \(\frac{π}{5}\)(c) θ = \(\frac{nπ}{7}\) + \(\frac{π}{3}\)(d) θ = \(\frac{nπ}{7}\) + \(\frac{π}{2}\)
Answer» Answer : (a) θ = \(\frac{nπ}{7}\) + \(\frac{π}{14}\) tan 5θ = cot 2θ ⇒ tan 5θ = tan (π/2 – 2θ) ⇒ 5θ = nπ + (π/2 – 2θ), n∈I ⇒ 7θ = nπ + π/2, n∈I ⇒ θ = \(\frac{nπ}{7}\) + \(\frac{π}{14}\) , n∈I.

The general solution of tan 5θ = cot 2θ is(a) θ = \(\frac{nπ}{7}\) + \(\frac{π}{14}\)(b) θ = \(\frac{nπ}{7}\) + \(\frac{π}{5}\)(c) θ = \(\frac{nπ}{7}\) + \(\frac{π}{3}\)(d) θ = \(\frac{nπ}{7}\) + \(\frac{π}{2}\)

Answer»

Answer : (a) θ = \(\frac{nπ}{7}\) + \(\frac{π}{14}\)

tan 5θ = cot 2θ

⇒ tan 5θ = tan (π/2 – 2θ)

⇒ 5θ = nπ + (π/2 – 2θ), n∈I

⇒ 7θ = nπ + π/2, n∈I

⇒ θ = \(\frac{nπ}{7}\) + \(\frac{π}{14}\) , n∈I.