tan! 13 -cot-(-V3) 36147(1) në B-TOUho(C) od

1.	tan! 13 -cot-(-V3) 36147(1) në B-TOUho(C) od
Answer» tan⁻¹√3 - cot⁻¹√3 = tan⁻¹(tan (π/3)) - cot⁻¹(-cot(π/6)) = tan⁻¹(tan (π/3)) - cot⁻¹(cot(π - π/6)) = (π/3) - (5π/6) = -π/2 AS we knowthat tan 60=underroot3so We know that tan¯¹ √3 is at the angle π/3.Now, ⇒ tan¯¹ √3 - cot-1 (-√3)⇒ π/3 - (π-π/3)⇒ π/3 - 5π/6⇒ -3π/6⇒ -π/2

Answer»

tan⁻¹√3 - cot⁻¹√3 = tan⁻¹(tan (π/3)) - cot⁻¹(-cot(π/6)) = tan⁻¹(tan (π/3)) - cot⁻¹(cot(π - π/6)) = (π/3) - (5π/6) = -π/2

AS we knowthat tan 60=underroot3so

We know that tan¯¹ √3 is at the angle π/3.Now,

⇒ tan¯¹ √3 - cot-1 (-√3)⇒ π/3 - (π-π/3)⇒ π/3 - 5π/6⇒ -3π/6⇒ -π/2

Discussion