Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev

The document Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

HOTS QUESTIONS

Q1. What is the maximum value of  Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev

Hint: ∵  Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev  = 3 sin θ
Since maximum value of ‘sin θ’ is 1
∴  maximum value of 3 sin θ is 3 × 1 i.e. 3

⇒  maximum value of Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev is 3.

Q2. If sin θ = 1/3,  then find the value of 9 cot2θ + 9.

Hint: cot2 θ = cosec2 θ – 1

Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev

∴ 9 cot2θ + 9 = 9(8) + 9 = 72 + 9 = 81

Q3. If 4 tan θ = 3, then find the value of  Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev

Hint: 

Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev

Q4. If sin α = 1/2 and cos β = 1/2 then find the value of (α + β).

Hint:

Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev

Q5. If sin θ + cos θ = √3 , find the value of tan θ + cot θ .

Hint: sin θ + cos θ = 3 ⇒ (sin θ + cos θ)2 = 3
⇒ sin2 θ + cos2 θ + 2 sin θ . cos θ = 3
⇒ 1 + 2 sin θ . cos θ = 3 ⇒ 2 sin θ . cos θ = 2    [∴ sin2θ + cos2θ = 1]

⇒ sin θ . cos θ =1 ⇒  1 =   Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev

⇒  Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev

Thus, tanθ + cotθ = 1

Q6. cos (A+B) = 1/2 and sin (A–B) = 1/2 ; 0° < (A + B) < 90° and (A – B) > 0°. What are the values of ∠A and ∠B?

Hint:

Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev    ...(1)

Hots Questions- Introduction to Trigonometry Class 10 Notes | EduRev    ...(2)

Adding (1) and (2), 2A = 90 ⇒ A = 45
From (1) 45° + B = 60°  ⇒   B = 60° – 45° = 15°
Thus, ∠A = 45° and ∠B = 15°

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