Trigonometry

1 )

If 2sec²α – sec⁴α – 2cosec²α + cosec⁴α = (15 / 4), then tan²α = _______

(1 / √2)

(1/2)

[1 / (2√2)]

(1/4)

View Solution

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).

2 )

If the roots of the quadratic equation x² + Ax + B = 0 are tan30° and tan15° then the value of A – B = ________

– 1

View Solution

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.

3 )

If θ = (6π / 7) and x = tanθ + cot(– θ) then

x > 0

x < 0

x = 0

x ≥ 0

View Solution

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 .

4 )

0 < A, B < (π/2) If tanA = (7/8), tanB = (1 / 15) then the value of A + B = ________

(π/2)

(π/3)

(π/4)

(π/6)

View Solution

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).

5 )

x + y = (π/2), then range of cosx ∙ cosy is

[– 1, 1]

[0, 1]

[– (1 / √2), (1 / √2)]

[– (1/2), (1/2)]

View Solution

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)].

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