Table of contents  
Previous Year Questions 2024  
Previous Year Questions 2023  
Previous Year Questions 2022  
Previous Year Questions 2021  
Previous Year Questions 2020  
Previous Year Questions 2019 
Q1: If sin α = √3/2 , cos β = √3/2 then tan α. tan β is: (2024)
(a) √3
(b) 1/√3
(c) 1
(d) 0
Ans: (c)
sin α = √3/2, ⇒ sin α = sin 60º
⇒ α = 60º
∵ cos β = √3/2,
⇒ cos β = cos 30º
⇒ β = 30º
tan α. tan β = tan 60º. tan 30º
= 1
Q2: Evaluate: (2024)
Ans:
Q3: Prove that: (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1 (2024)
Ans:
L.H.S. = (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ)
= (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ)
= 1 = R.H.S.
Hence, proved.
Q4: If 2 tan A = 3, then find the value of is (2023)
View AnswerAns:
Hence, the answer is 1.
Q5: [5/8 sec^{2}60°  tan^{2}60° + cos^{2}45° is equal to (2023)
(a) 5/3
(b) 1/2
(c) 0
(d) 1/4
Ans: (c)
Sol:
Q6: Evaluate 2 sec^{2}θ + 3 cosec^{2}θ  2 sin θ cos θ if θ = 45° (2023)
Ans: Since θ = 45°, sec 45° = √2, cosec 45° = √2, sin 45°^{ }= 1/√2 cos 45° = 1/√2
2sec^{2} θ + 3 cosec^{2} θ – 2 sin θ cos θ
= 4 + 6 – 1 = 9
Q7: Which of the following is true for all values of θ(0^{o} ≤ θ ≤ 90^{o})? (2023)
(a) cos^{2}θ  sin^{2}θ  1
(b) cosec^{2}θ  sec^{2}θ 1
(c) sec^{2}θ  tan^{2}θ  1
(d) cot^{2}θ tan^{2}θ = 1
Ans: (c)
Q8: If sinθ +cosθ = √3. then find the value of sinθ . cosθ. (2023)
Ans: Given, sinθ +cosθ = √3
Squaring both sides, we get (sinθ + corsθ)^{2} = 3
⇒ sin^{2}θ + cos^{2}θ + 2sinθ cosθ = 3
⇒ 2sinθ cosθ = 3  1 ( ∵ sin^{2}θ + cos^{2}θ = 1)
⇒ 2sinθ cosθ = 2
⇒ sinθ cosθ = 1
Q9: If sin α = 1/√2 and cot β = √3, then find the value of cosec α + cosec β. (2023)
Ans: Given, sin α = 1/√2 and cot β = √3
We know that, cosec α = 1/sinα = √2
Also, 1 + cot^{2}β = cosec^{2}β
⇒ cosec^{2}β = 4
⇒ cosec β = 4
Now, cosec α + cosec β = √2 + 2
Q10: Prove that the Following Identities: Sec A (1 + Sin A) ( Sec A  tan A) = 1 (2023)
Ans: LHS = sec A(1 + sin A )( sec A  tan A)
= 1
= RHS
Hence proved..
Ans: (c)
Sol:
Given, cosθ = √3/2 = B/H
Let B = √3k and H = 2k
∴ [By Pythagoras Theorem]
⇒√k^{2} = k
Q12: is equal to (2022)
(a) 0
(b) 1
(c) sinθ + cosθ
(d) sinθ  cosθ
Ans: (c)
Sol: We have,
Q13: The value of θ for which 2 sin2θ = 1, is (2022)
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Ans: (a)
Sol: Given, 2 sin2θ = 1 ⇒ sin2θ = 1/2
⇒ 2θ = 30°
⇒ θ = 15°
Q14: If sin^{2}θ + sinθ = 1, then find the value of cos^{2}θ + cos^{4}θ is (2022)
(a) 1
(b) 1
(c) 0
(d) 2
Ans: (b)
Sol: Given, sin^{2}θ + sinθ = 1 (i)
sinθ = 1  sin^{2}θ
⇒ sinθ = cos^{2}θ (ii)
∴ cos^{2}θ + cos^{4}θ
= sinθ + sin^{2}θ [From (ii)]
= 1 [From (i)]
Q15: If 3 sin A = 1. then find the value of sec A. (2021 C)
View AnswerAns: We have, 3 sin A = 1
∴ sin A = 1/3
Now by using cos^{2 }A = 1  sin^{2} A, we get
Q16: Show that: (2021 C)
Ans: We have, L.H.S.
[By using 1 + tan^{2}θ = sec^{2}θ and 1 + cot^{2} θ = cosec^{2}θ ]
Hence,
Ans: (a)
Sol: We have, sin θ = cos θ
or sin θ / cos θ = 1
⇒ tan θ = 1 and cot θ = 1 [∵ cot θ = 1/tanθ]
∴ tan^{2 }θ + cot^{2 }θ = 1^{2 } + 1^{2 } = 2
Hence, A option is correct.
Q18: Given 15 cot A = 8, then find the values of sin A and sec A. (2020)
Ans: In right angle ΔABC we have
15 cot A = 8
⇒ cot A = 8/15
Since, cot A = AB/BC
∴ AB/BC = 8/15
Let AB = 8k and BC = 15k
By using Pythagoras theorem, we get
AC^{2 }= AB^{2} + BC^{2}
⇒ (8k)^{2} + (15)^{2} = 64k^{2} + 225k^{2} = 289k^{2} = (17k)^{2 }
So, sec A = 1/cosA = 17/8
Q19: Write the value of sin^{2} 30° + cos^{2} 60°. (2020)
Ans: We have, sin^{2} 30° + cos^{2} 60°
Q20: The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ − b cos θ) is (2020)
(a) a^{2 }+ b^{2}
(b) a + b
(c)
(d)
Ans: (c)
Sol: Given the point A (cos θ + b sin θ , 0), (0 , a sin θ − b cos θ)
By distance formula,
The distance of
[∵ cos^{2}θ + sin^{2}θ = 1]
Q21: 5 tan^{2}θ  5 sec^{2}θ = ____________. (2020)
Ans: We have 5(tan^{2}θ  sec^{2}θ)
= 5(1) =  5 [By using 1 + tan^{2}θ = sec^{2} θ ⇒ tan^{2}θ  sec^{2}θ =  1]
Q22: If sinθ + cosθ = √3. then prove that tan θ + cot θ = 1 (2020)
Ans: sin θ + cos θ =√3
= (sinθ + cosθ)^{2 }= 3
= sin^{2} θ + cos^{2} θ + 2sin θ cos θ = 3
⇒ 2sin θ cos θ = 2
⇒ sin θ cos θ = 1
⇒ sin θ cos θ = sin^{2}θ + cos^{2}θ
⇒ tan θ + cot θ = 1
Q23: If sin x + cos y = 1, x = 30° and y is acute angle, find the value of y. (2019)
View AnswerAns: Given,
⇒ sin x + cos y = 1
⇒ sin 30° + cos y = 1
⇒ 1/2 + cos y = 1
⇒ cos y = 1  1/2
⇒ cos y = 1/2
⇒ cos y = cos 60°.
Hence, y = 60°.
Q24: If cosec^{2} θ (cos θ  1)(1 + cos θ) = k, then what is the value of k? (2019)
Ans: Given:
cosec2 θ (cos θ  1)(1 + cos θ) = k
Concept used:
Cosec α = 1/Sin α
Sin^{2} α + Cos^{2} α = 1
(a + b)(a  b) = a^{2}  b^{2}
Calculation:
cosec2 θ (cos θ  1)(1 + cos θ) = k
⇒ cosec2 θ (1  cos θ)(1 + cos θ) = k
⇒ cosec2 θ (1  cos^{2} θ) = k
⇒ cosec2 θ × sin^{2} θ = k
⇒ 1 = k
⇒ k = 1
∴ The value of k is (1).
Q25: The value of ( 1 + cot A − cosec A ) ( 1 + tan A + sec A ) is
Ans:
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