If the equation tan θ + tan 2θ + tan θ . tan 2θ = 1, θ =(a) \(\f

1.	If the equation tan θ + tan 2θ + tan θ . tan 2θ = 1, θ =(a) \(\frac{nπ}{6}\) +\(\frac{π}{6}\)(b) \(\frac{nπ}{2}\) + 6(c) \(\frac{nπ}{3}\)+ \(\frac{π}{12}\)(d) \(\frac{nπ}{2}\)+\(\frac{π}{12}\)
Answer» Answer : (c) \(\frac{n\pi}{3}\)+ \(\frac{\pi}{12}\) tan θ + tan 2θ = 1 – tan θ . tan 2θ ⇒ \(\frac{tan\,θ +tan\,2θ}{1- tan\,θ\,tan\,2θ}\) =1 ⇒ tan (θ + 2θ) = 1 \(\big( \because tan\,(A+B) = \frac{tan\,A+tan\,B}{1-tan\,A\,tan\,B}\big)\) ⇒ tan 3θ = tan \(\frac{\pi}{4}\) ⇒ 3θ = nπ + \(\frac{\pi}{4}\) ⇒ θ = \(\frac{n\pi}{3}\)+ \(\frac{\pi}{12}\).

If the equation tan θ + tan 2θ + tan θ . tan 2θ = 1, θ =(a) \(\frac{nπ}{6}\) +\(\frac{π}{6}\)(b) \(\frac{nπ}{2}\) + 6(c) \(\frac{nπ}{3}\)+ \(\frac{π}{12}\)(d) \(\frac{nπ}{2}\)+\(\frac{π}{12}\)

Answer»

Answer : (c) \(\frac{n\pi}{3}\)+ \(\frac{\pi}{12}\)

tan θ + tan 2θ = 1 – tan θ . tan 2θ

⇒ \(\frac{tan\,θ +tan\,2θ}{1- tan\,θ\,tan\,2θ}\) =1

⇒ tan (θ + 2θ) = 1 \(\big( \because tan\,(A+B) = \frac{tan\,A+tan\,B}{1-tan\,A\,tan\,B}\big)\)

⇒ tan 3θ = tan \(\frac{\pi}{4}\)

⇒ 3θ = nπ + \(\frac{\pi}{4}\)

⇒ θ = \(\frac{n\pi}{3}\)+ \(\frac{\pi}{12}\).