HirePro Trigonometry Quiz-1

Here, cot A + cosec A = 3

⇒ (cos A / sin A) + (1 / sin A) = 3
⇒ cos A + 1 = 3sin A

Squaring both sides.

⇒ (cos A + 1)2 = 9 Sin2A
⇒ cos2 A + 2 Cos A + 1 = 9 (1 – cos2 A)
⇒ 10 cos2A + 2 cos A – 8 = 0
⇒ 5 cos2 A + cosA – 4 = 0
⇒ (5 cos A – 4)(cos A + 1) = 0
(∵ A is an acute angle)
∴ cos A = (4 / 5)

sin(45° + θ) – cos(45° – θ) = sin{90° – (45° – θ)} – cos (45° – θ)
= cos (45° – θ) – cos (45° – θ) {∵sin(90 – A) = cos A }
= 0

Dividing numerator and denominator by sin θ
⇒ [{(2sin θ – cos θ) / Sin θ} / {(cos θ + sin θ) / sin θ}] = 1
⇒ {(2 – cot θ) / (1 + cot θ)} = 1
⇒ 2 cot θ = 1 + cot θ
⇒ cot θ = 1 / 2

Two angles = A and B where A > B ∴ A + B = 135°

= \frac{(135 \times \prod)}{180} radian
⇒ A + B = (3 \prod / 4) radian .....(i)
⇒ A – B = (\prod / 12) radian .....(ii)

on adding these equations,

2A = (3\prod / 4) + (\prod / 12)
= 9\prod + \prod / 12 = 10\prod / 12 = 5\prod / 6

∴ A = 5\prod / 12 radian

2 sin^{2}\Theta + 3 cos^{2}\Theta
= 2 Sin ^{2}\Theta + 2 Cos^{2}\Theta + Cos ^{2}\Theta
= 2 (Sin ^{2}\Theta +Cos ^{2}\Theta) + Cos ^{2}\Theta
= 2 + cos^{2}\Theta

[ ∵ Sin ^{2}\Theta+ Cos ^{2}\Theta= 1 ]

∵ Minimum value of cos θ = –1
But cos^{2}\Theta ≥ 0, when θ = 90°

[ ∵ cos 0° = 1, cos 90° = 0 ]
∴ Required minimum value = 2 + 0 = 2.

cos x + cos y = 2
∵ cos x ≤ 1

⇒ cos x = 1 ; cos y = 1
⇒ x = y = 0° [∵ cos 0° = 1]
∴ sin x + sin y = 0



 

Let the measure of three angles of triangle are 2x, 7x and 11x respectively.

∴ 2x + 7x + 11x = 180°
⇒ 20x = 180°
⇒ x = 9°
∴ First angle = 2x = 2 × 9° = 18°
Second angle = 7x = 7 × 9° = 63°
Third angle = 11x = 11 × 9° = 99°

 

tan 4° . tan 43° . tan 47° . tan 86°

= tan 4° . tan 43° . tan (90° – 43°) . tan (90° – 4°)
= tan 4° . tan 43° . cot 43° . cot 4°
[∵ tan (90° – Θ) = cot Θ & tan Θ . cot Θ = 1]
= 1

tan 28° = (tan 90° - 62°)
cot 62º = q / p

Sin 3A = cos (A – 26°)
⇒ cos (90° – 3A) = cos (A – 26°)

[∵ cos (90° – Θ) = sin Θ]

⇒ 90° – 3A = A – 26°
⇒ 4A = 116°
⇒ A = 29°