Question

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 1

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.