Trigonometric EquationHard
Question
Let cos(α + β) =
and let sin(α - β) =
, where 0 ≤ α, β ≤
, then tan 2α =
and let sin(α - β) =
, where 0 ≤ α, β ≤
, then tan 2α = Options
A.

B.

C.

D.

Solution
cos (α + β) =
⇒ tan( α + β) =
sin (α - β) =
⇒ tan( α - β) =
tan 2α = tan(α + β + α - β) =
⇒ tan( α + β) =
sin (α - β) =
⇒ tan( α - β) =
tan 2α = tan(α + β + α - β) =

Create a free account to view solution
View Solution FreeMore Trigonometric Equation Questions
If tan θ - √2 sec θ = √3, then the general solution of θ -...The general solution of tan 3x = 1 is -...The value of ∫-π2π2sin2x1+2x dx is :...Total number of solution of 16cos2x + 16sin2x = 10 in x ∈ [0, 3π] is equal to -...The general value of θ satisfying sin2 θ + sin = 2 is -...