If tanx + tan x + π 3 + tan x + 2 π 3 = 3 , then prove that 3 tanx - tan 3 x 1 - 3 tan 2 x = 1

To prove: 3 tanx - tan 3 x 1 - 3 tan 2 x = 1 Consider, tanx + tan x + π 3 + tan x + 2 π 3 = 3 ⇒ tanx + tanx + tan π 3 1 - tanx tan π 3 + tanx + tan 2 π 3 1 - tanx tan 2 π 3 = 3 ∵ tan A + B = tanA + tanB 1 - tanA tanB ⇒ tanx + tanx + 3 1 - 3 tanx + tanx - 3 1 + 3 tanx = 3 ∵ tan π 3 = 3 and tan 2 π 3 = - 3 ⇒ tanx ( 1 - 3 tan 2 x ) + ( tanx + 3 ) ( 1 + 3 tanx ) + ( tanx - 3 ) ( 1 - 3 tanx ) 1 - 3 tan 2 x = 3 ⇒ tanx - 3 tan 3 x + tanx + 3 tan 2 x + 3 + 3 tanx + tanx - 3 tan 2 x - 3 + 3 tanx 1 - 3 tan 2 x = 3 ⇒ 9 tanx - 3 tan 3 x 1 - 3 tan 2 x = 3 ⇒ 3 tanx - tan 3 x 1 - 3 tan 2 x = 1 Hence proved.

If a right angle be divided into three parts α , β and γ , prove that cot α = tan β + tan γ 1 − tan β tan γ

Given, α + β + γ = π 2 ⇒ β + γ = π 2 - α Taking tan of both sides tan ( β + γ ) = tan ( π 2 − α ) As tan π 2 - α lies in first quadrant and here it is positive. ⇒ tan β + tan γ 1 − tan β tan γ = cot α .

If sin α + β = 1 and sin α - β = 1 2 , where 0 ≤ α , β ≤ π 2 , then find the value of tan α + 2 β and tan 2 α + β .

Given, 0 < α , β ≤ π 2 sin α + β = 1 ⇒ α + β = π 2 ...(i) And sin α - β = 1 2 ⇒ α - β = π 6 ...(ii) From (i) and (ii), α = π 3 and β = π 6 Now, tan α + 2 β = tan π 3 + 2 × π 6 = tan 2 π 3 = tan π - π 3 = - tan π 3 = - 3 And tan ( 2 α + β ) = tan ( 2 × π 3 + π 6 ) = tan 5 π 6 = tan ( π - π 6 ) = - tan π 6 = - 1 3

If m tan θ - 30 ° = n tan θ + 120 ° , show that, cos 2 θ = m + n 2 m - n .

Given: m tan θ - 30 ° = n tan θ + 120 ° and we have to prove cos 2 θ = m + n 2 ( m - n ) . We can write given relation as: m n = tan θ + 120 ° tan θ - 30 ° = - cot 30 ° + θ tan θ - 30 ° Here, now take - sign common from angle of denominator and cancel with - sign of numerator. ∴ m n = cot 30 ° + θ tan 30 ° - θ m n = cos 30 ° + θ · cos 30 ° - θ sin 30 ° + θ · sin 30 ° - θ Now, using sin A + B sin A - B = sin 2 A - sin 2 B and cos A + B cos A - B = cos 2 A - sin 2 B , we get, m n = cos 2 30 - sin 2 θ sin 2 30 ° - sin 2 θ = 3 4 - sin 2 θ 1 4 - sin 2 θ Now, apply componendo and dividendo. ⇒ m + n m - n = 1 - 2 sin 2 θ 1 2 Now, using cos 2 A = 1 - 2 sin 2 A ⇒ m + n m - n = 2 cos 2 θ ∴ cos 2 θ = m + n 2 ( m - n )

If α + β = γ and tan α tan β = x y , then prove that sin ( α - β ) = x - y x + y sin γ .

Given that, tan α tan β = x y Apply componendo and dividendo i.e., a b = c d implies a ± b a ∓ b = c ± d c ∓ d . = tan α + tan β tan α - tan β = x + y x - y ⇒ sin α cos α + sin β cos β sin α cos α - sin β cos β = x + y x - y ⇒ sin α cos β + sin β cos α sin α cos β - sin β cos α = x + y x - y ⇒ sin ( α + β ) sin ( α - β ) = x + y x - y ⇒ sin ( α - β ) = x - y x + y sin α + β ⇒ sin ( α - β ) = x - y x + y sin γ Hence, proved that sin ( α - β ) = x - y x + y sin γ .