Trigonometric EquationHard
Question
If tan2 θ - (1 + √3) tan θ + √3 = 0, then the general value of θ is :
Options
A.nπ + 
, nπ + 
, nπ + B.nπ - , nπ +
, nπ + C.nπ + , nπ -
, nπ - D.nπ - , nπ -
, nπ - Solution
tan2 θ - tan θ - √3 [tan θ + √3 = 0
tan θ [tan θ -1] - √3 [tan θ -1] = 0
⇒ tan θ = 1, tan θ = √3
θ = nπ +
, θ = nπ + 
tan θ [tan θ -1] - √3 [tan θ -1] = 0
⇒ tan θ = 1, tan θ = √3
θ = nπ +
, θ = nπ + Create a free account to view solution
View Solution FreeMore Trigonometric Equation Questions
equals...The volume of the tetrahedron formed by the coterminous edges Then the volume of the parallelopiped formed by the coterm...tan A + cot(180o + A) + cot (90o + A ) + cot (360o - A) =...If one of the lines of my2 + (1 - m2)xy - mx2 = 0 is a bisector of the angle between the lines xy = 0, then m is...In a triangle ABC, ∠B = and ∠C = LetDdivides BC internally in the ration 1 : 3, then is equal to...