Trigonometry

Answer :

given : 2sec²α – sec⁴α – 2cosec²α + cosec⁴α = (15 / 4)

                   consider LHS = 2(1 + tan²α) – (1 + tan²α)² – 2(1 + cot²α) + (1 + cot²α)²

                                       = 2[1 + tan²α – 1 – cot²α] – (1 + 2tan²α + tan⁴α)

                                                                                      + (1 + 2cot²α + cot⁴α)

                                       = 2(tan²α – cot²α) – 2tan²α – tan⁴α + 2cot²α + cot⁴α

                                       = cot⁴α – tan⁴α

                                       = [1 / (tan⁴α)] – tan⁴α

                   ∴ [{1 – (tan⁸α)} / (tan⁴α)] = (15 / 4)

                   Let tan⁴α = P

                   hence [(1 – P²) / P] = (15 / 4)

                   ∴ 4P² + 15P – 4 = 0

                   hence P = – 4 or (1/4)

                   i.e. tan⁴α = (1/4)

                   hence tanα = (1 / √2).

Answer :

given: roots of x² + Ax + B are tan30, tan15.

                   ∴ tan30 + tan15 = – A

                             tan30 ∙ tan15 = B

                   As tan(α + β) = [(tanα + tanβ) / (1 – tanα ∙ tanβ)]

                   ∴ tan45° = [(tan30 + tan15) / (1 – tan30 ∙ tan15)]

                             1 = [(– A) / (1 – B)]

                   ∴ 1 – B = – A

                   ∴ A – B = – 1.

Answer :

given: θ = (6π / 7)

                   also x          = tanA + cot(– A)

                             = tanA – cotA

                             = [(sin²θ – cos²θ) / (sinθ ∙ cosθ)]

                             = [(– 2cos2θ) / (sin2θ)]

                             = – 2 cot2θ

                   as θ = (6π / 7) ⇒ x = (– 2)cot(12π / 7) ⇒ x = (– 2) × some negative number

                   ⇒ x > 0 .

Answer :

0 < A, B < (π/2)

                   tanA = (7/8), tanB = (1 / 15)

                   ∵ tan(A + B) = [(tanA + tanB) / (1 – tanA ∙ tanB)]

                    ∴ tan(A + B)        = [{(7/8) + (1 / 15)} / {1 – [(7/8) × (1 / 15)]}]

                                      = [{(113) / (120)} / {1 – [7 / (120)]}]

                                      = 1

                   ∴ A + B = tan^–1(1) = (π/4).

Answer :

x + y = (π/2) ⇒ y = (π/2) – x

                    cosx ∙ cosy = cosx ∙ cos[(π/2) – x]

                                      = cosx ∙ sinx

                                      = [(sin2x) / 2]

                   ∵ sinx ∈ [– 1, 1]

                   ⇒ [(sin2x) / 2] ∈ [– (1/2), (1/2)].