Previous Year Questions: Introduction to Trigonometry - Mathematics (Maths) Class 10

Short Answer Type Questions

Q1: sin² 60° + 2 tan 45° - cos² 30°    [Allahabad 2019] 
Ans:  2


Q2: If sin x + cos y = 1; x = 30° and is an acute angle, find the value of y.  [CBSE 2019 (30/5/1)]
Ans: We have, sin x + cosy = 1
⇒ sin30° + cosy = 1


⇒ y = 60⁰

Q3: Prove that: (1 + cot A - cosec A)(l + tan A + sec A) = 2   [CBSE 2019 (30/1/2)]
Ans: LHS = (1 + cotA-cosecA) (1 + tanA +secA)




Q4: Prove that:        [CBSE 2019 (30/5/1)]
Ans: 


Q 5: If cos θ + sin θ = √2cos θ , show that cos θ - sin θ = √2 sin θ.  [CBSE 2019 (30/ 5/ 1)]
Ans: Given: cos θ + sin θ = √2cos θ

Now, (cos θ + sin θ)² + ( cos θ - sin θ)² 
cos²θ + sin²θ + 2cosθ. sin θ + cos²θ + sin²θ - 2cosθ. sinθ
1 + 1 = 2

Again, (cos θ + sin θ)² + (cos θ - sin θ)² = 2
⇒ (√2 cosθ)² + (cos θ - sin θ)² = 2
⇒ 2 cos²θ + ( cos θ - sin θ)² = 2
 ⇒ (cos θ - sin θ)² = 2 - 2cos²θ = 2 ( 1 - cos²θ)
= 2sin²θ
∴ cos θ - sin θ = √2 sin θ
Hence proved

Q6: Prove that: ( sin A + cosec A)² + ( cos A + sec A)² = 7 + tan² A + cot² 4    

[CBSE 2019 (30/1/2;]

Ans: LHS = (sin A + cosec A)² + (cos A + sec A)² 
=sin² A + cosec² A + 2sinA . cosec A + cos² A + sec² A + 2 cos A .sec A
= (sin² A + cosec² A + 2) + (cos² A + sec² A + 2)     (sin A.cosec A = 1)
= (sin² A + cos²A) + (cosec² A + sec² A) + 4      (cos A. sec A = 1)
= 1 + 1+ cot² A + 1 + tan² A + 4
= 7 + tan² A + cot² A = RHS (∵ 1 + cot² A = cosec² A and 1 + tan² A = sec² A)

Q7: If sin (A + 2B)   and cos (A + 4B) = 0, A > B, and A + 4B ≤ 90⁰ then find A and B. [CBSE 2018 (C)]
Ans: 
So, sin(A + 2B) = sin 60°
Hence A + 2B = 60°     ..(i)
Also, we have
cos(A + 4B) = 0
cos(A + 4B) = cos 90°
A + 4B = 90°     ..(ii)
Subtracting (ii) from (i), we have
-2B = - 30°
=> B = 15°
Put B = 15° in eq. (i),  we have
A + 2(15°) = 60°
⇒ A + 30° = 60°
⇒ A = 30°

Long Answer Type Questions

Q1: Prove that:  [CBSE 2019 (30/5/1)]
Ans: LHS

Q2: Prove that:     [CBSE 2019 (30/5/1)]
Ans: LHS