SSC CGL Previous Year Questions: Trigonometry - 5 - SSC CGL MCQ

In ΔABC
tan α = h/9 ... (1)
In ΔABD
tan β = h/16

α + β = 90° (given)
β = 90 – α
Since, tan β = h/16

From eqn. (1) and (2)

Dividing Numerator and Denominator by cos θ

AB = AC
∴ ∠ABC = ∠ACB ...(1)

[opposite angles of equal sides are equal]
AC = AD and BA = AC
∴ ∠ACD = ∠ADC and ∠ABC = ∠ACB ...(2)
In a triangle,
∠ABC + ∠ADC + ∠DCB = 180° (Angle sum property)
∠ABC + ∠ADC + ∠ACB + ∠ACD = 180°
2∠ACB + 2∠ACD = 180°
[From eqn. (1) & (2)]
∴ ∠BCD = 90° or π / 2

sec² θ – tan² θ = 1
(sec θ + tan θ) (sec θ – tan θ) = 1

2y cos θ = x sin θ

sinθ + sin²θ = 1
⇒ sinθ = 1 – Sin²θ
⇒ sinθ = cos2θ
∴ cos¹²θ + 3cos¹⁰θ + 3cos⁸θ + cos⁶θ – 1 
= (cos⁴θ + cos²θ)³ – 1
= (sin²θ + cos²θ)³ – 1 = 1 – 1 = 0

AB = Pole = h metre
BD = x metre
From ΔABC,

2y cos θ = xsin θ
⇒ x sec θ = 2y cosec θ
Also, 2x sec θ – y cosec θ = 3
⇒ 4y cosec θ – y cosec θ = 3
⇒ 3y cosec θ = 3
⇒ y cosec θ = 1
⇒ y = sin θ
∴ x sec θ = 2y cosec θ
= 2sin θ . cosec θ = 2
⇒ x = 2cos θ
∴ x² + 4y² = 4cos²θ + 4sin²θ = 4

tan (x + y). tan (x – y) = 1
⇒ tan (x + y) = cot (x – y) = tan (90°– x + y)
⇒ x + y = 90° – x + y ⇒ 2x = 90°

4 sec² θ + 9cosec² θ
= 4 (1 + tan² θ) + 9 (1 + cot² θ)
= 4 + 4 tan² θ + 9 + 9cot² θ
= 4 tan² θ + 9cot²θ +  12 – 12 + 13
= (2tan² θ – 3Cot² θ)² + 25
{∵ least value of 2 tan² θ – 3cot² θ = 0}
∴ the minimum value is 25.

sin (2x – 20°) = cos (2y + 20°)
⇒ sin (2x – 20°) = sin (90° – 2y – 20°) = sin (70° – 2y)
⇒ 2x – 20° = 70° – 2y ⇒ 2 (x + y) = 90° ⇒ x + y = 45°
∴ sec (x + y) = sec 45° = √2

2sinα + 15cos²α = 7
⇒ 2 sinα + 15 (1 – sin²α) = 7
⇒ 2 sinα + 15 – 15 sin²α = 7
⇒ 15 sin²α – 2 sinα – 8 = 0
⇒ 15 sin²α – 12 sinα + 10 sinα – 8 = 0
⇒ 3 sinα (5 sinα – 4) + 2 (5 sinα – 4) = 0
⇒ (3 sinα + 2) ( 5 sinα – 4) = 0
⇒ 5 sinα – 4 = 0
⇒  sinα = 4/5
∴ cosecα = 5/4

Put θ = 45°
a = tan 45° – cot 45°,  b = sin 45° – cos 45°
a = 1 – 1 = 0

Put in equation
(a² + 4) (b² –1)² = (0 + 4) (0 – 1)² = 4

(a – b)³ = a³ – b³ – 3ab (a – b)
⇒ 8 = 56 – 3ab (2)
⇒ 6ab = 56 – 8 = 48
⇒ 2ab = 16 ...(i)
∴ a² + b² = (a – b)² + 2ab
= 4 + 16 = 20

4 cosec²α + 9sin²α
= 4 cosec²α + 4 sin²α + 5 sin²α
= 4 [(cosecα – sinα)² + 2] + 5 sin²α
= 12  [∵ cosecα – sinα ≥ 1]

tan (x + y) tan (x – y) = 1
⇒ tan (x + y) = cot (x – y) = tan (90° – (x – y))
⇒ x + y = 90°  – ( x – y)
⇒ 2x = 90°
⇒  x = 45°
∴ tan x = tan 45° = 1

Let BC = x
∴ AB = x + 2
∴ AB² + BC² = AC²
⇒ (x + 2)² + x² = (2√5) ² ⇒ x² + 4x + 4 + x² = 20
⇒ 2x² + 4x – 16 = 0
⇒ x² + 2x – 8 = 0 ⇒ x² + 4x – 2x – 8 = 0
⇒ x (x + 4)  – 2 (x + 4) = 0
⇒ (x – 2) (x + 4) = 0
⇒ x = 2 = BC
∴ AB = 2 + 2 = 4 cm
cos A = AB/AC = 4/2√5 = 2/√5
cos C = BC/AC = 2/2√5 = 1/√5
(cos²A – cos²C) = (2/√5)²  -  (1/√5)² = 4/5 - 1/ 5 = 3/5

cot A +  cosec A = 3
cosec² A – cot² A = 1
(cosec A – cot A) (cosec A + cot A) = 1
cosec A – cot A = 1/3 ... (i)
cosec A + cot A = 3 ... (ii)
By Adding (i) and (ii),

cos A = 1 – cos² A = sin2A
∴ sin²A + sin⁴A = sin²A + cos²A = 1

AB = tree
BC = broken part
∴ BC = CD
AD = 30 metre


∴ AB = AC + BC
= 10√3 + 20√3 = 30√3 metre

sec⁴θ – sec²θ = sec² θ (sec²θ – 1)
= (1 + tan²θ) (1 + tan²θ – 1) = tan²θ + tan⁴θ

x sin³θ + y cos³θ = sinθ.cosθ
=> (x sinθ). sin²θ + (y cosθ) cos²θ = sinθ.cosθ
=> xsinθ . sin²θ + xsinθ . cos²θ = sinθ . cosθ
=> x sinθ (sin²θ + cos²θ) = sinθ . cosθ
=> x = cosθ
now, x sinθ = y cosθ
=> cosθ . sinθ = y cosθ
=> y = sinθ
x² + y² = cos²θ + sin²θ = 1