The most general solution of tan θ = –1 and cos θ = \(\frac{1}{\sq

1.	The most general solution of tan θ = –1 and cos θ = \(\frac{1}{\sqrt{2}}\) is (a) nπ + (–1)n \(\frac{π}{4}\) (b) 2nπ + \(\frac{3π}{4}\)(c) nπ + (–1)n \(\frac{5π}{4}\) (d) 2nπ + \(\frac{7π}{4}\)
Answer» Answer : (d) 2nπ + \(\frac{7π}{4}\) tan θ = –1 ⇒ tan θ = – tan π/4 = tan (π – π/4) = tan (2π - π/4) ⇒ tan θ = tan \(\frac{3π}{4}\) or tan \(\frac{7π}{4}\) ⇒ θ = \(\frac{3π}{4}\) , \(\frac{7π}{4}\) ...(i) cos θ = \(\frac{1}{\sqrt{2}}\) ⇒ cos θ = cos \(\frac{π}{4}\) = cos (2π – π/4) ⇒ cos θ = cos \(\frac{π}{4}\) = cos \(\frac{7π}{4}\) ⇒ θ = \(\frac{π}{4}\) , \(\frac{7π}{4}\)...(ii) ∴ From (i) and (ii) θ = \(\frac{7π}{4}\) is the only value of θ in [0, 2π] which satisfies both the equations. ∴ The general value of θ satisfying both the equations is θ = 2nπ + \(\frac{7π}{4}\)

The most general solution of tan θ = –1 and cos θ = \(\frac{1}{\sqrt{2}}\) is (a) nπ + (–1)n \(\frac{π}{4}\) (b) 2nπ + \(\frac{3π}{4}\)(c) nπ + (–1)n \(\frac{5π}{4}\) (d) 2nπ + \(\frac{7π}{4}\)

Answer»

Answer : (d) 2nπ + \(\frac{7π}{4}\)

tan θ = –1

⇒ tan θ = – tan π/4 = tan (π – π/4) = tan (2π - π/4)

⇒ tan θ = tan \(\frac{3π}{4}\) or tan \(\frac{7π}{4}\) ⇒ θ = \(\frac{3π}{4}\) , \(\frac{7π}{4}\) ...(i)

cos θ = \(\frac{1}{\sqrt{2}}\)

⇒ cos θ = cos \(\frac{π}{4}\)

= cos (2π – π/4)

⇒ cos θ = cos \(\frac{π}{4}\) = cos \(\frac{7π}{4}\) ⇒ θ = \(\frac{π}{4}\) , \(\frac{7π}{4}\)...(ii)

∴ From (i) and (ii) θ = \(\frac{7π}{4}\) is the only value of θ in [0, 2π] which satisfies both the equations.

∴ The general value of θ satisfying both the equations is

θ = 2nπ + \(\frac{7π}{4}\)