Previous Year Questions 2025

Q1: If tan A+ cot A= 6, then find the value of tan²A + cot² A – 4.

Ans: We have, tanA + cotA = 6
On squaring both sides, we get (tanA + cotA)² = 36
⇒ tan²A + cot²A + 2 tanA cotA = 36
Since tan A cot A = 1
⇒ tan²A + cot²A + 2 = 36
⇒ tan ²A + cot ²A = 36 – 2 = 34
∴ tan²A + cot²A – 4 = 34 – 4 = 30

Q2: If tan 3θ = √3, then θ/2 equals
(a) 60°
(b) 30°
(c) 20°
(d) 10°

–

Ans: (d)
We have, tan3θ = √3
⇒ tan3θ = tan 60°

Q3: If sin 4θ = √3/2, then θ/3 equals:
(a) 60°
(b) 20°
(c) 15°
(d) 5°

–

Ans: (d)
Given, sin 4θ = √3/2 = sin 60°.
⇒ 4θ = 60°
⇒ θ = 15°.

Q4: If α + β = 90° and α = 2β, then cos²α + sin²β is equal to:
(a) 0
(b) 1/2
(c) 1
(d) 2

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Ans: (b)
α + β = 90° and α = 2β …(i)
∴ 2β + β = 90° ⇒ 3β = 90° ⇒ β = 30°
∴ α = 2 × 30° = 60° [From (i)]
∴ cos²α + sin²β = cos²60° + sin²30°

Q5: then x : y =
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 4: 1

–

Ans: (c)

Q6: If 4k = tan²60° – 2cosec² 30° – 2tan² 30°, then find the value of k.

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Ans: Given, 4k = tan²60° – 2 cosec²30° – 2 tan²30°

Q7: If x cos60° + ycos0° + sin30° – cot45° = 5, then find the value of x + 2y.

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Ans: We have, x cos60° + y cos0° + sin30° – cot45° = 5

Q8:

–

Ans:

Q9:
(a) cot θ
(b)
(c)
(d) tan θ

–

Ans: (a)
We have,
[∵ 1 – cos²θ = sin²θ]

Q10: The value of (tan A cosec A)² – (sin A sec A)² is:
(a) 0
(b) 1
(c) -1
(d) 2

–

Ans: (b)
We have,
(tan A · cosec A)² – (sin A . sec A)²
Using identities:
tan A = sin A / cos A,
cosec A = 1 / sin A,
sec A = 1 / cos A.
sec² A – tan² A = 1

Q11: (cotθ + tanθ) equals:
(a) cosecθ secθ
(b) sinθ secθ
(c) cosθ tanθ
(d) sinθ cosθ

–

Ans: (a)
We have, cotθ + tanθ
cot θ = cos θ / sin θ,
tan θ = sin θ / cos θ.

Q12: The value of
(a) 1
(b) 0
(c) -1
(d) 2

–

Ans: (c)

Q13: In a right triangle ABC, right-angled at A, if sin B = 1/4 then the value of sec B is
(a) 4
(b) √15/4
(c) √15
(d) 4/√15

–

Ans: (d)
Given, sin B = 1/4

Q14: If a secθ + b tan θ = m and b sec θ + a tan θ = n, prove that a² + n² = b² + m²

–

Ans:
We have, a sec θ + b tan θ = m
and b sec θ + a tan θ = n
Taking, m² – n²
= (a sec θ + b tan θ)² – (b sec θ + a tan θ)²
= a²sec²θ + b²tan²θ + 2ab tan θ sec θ – b²sec²θ – a²tan²θ – 2ab tan θ sec θ
= a²(sec²θ – tan²θ) + b²(tan²θ – sec²θ)
= a² × 1 + b² × (-1)
= a² – b²
∴ m² – n² = a² – b²
⇒ a² + n² = b² + m²
Hence, proved.

Q15: Use the identity: sin²A + cos²A = 1 to prove that tan²A + 1 = sec²A. Hence, find the value of tan A, where sec A = 5/3, where A is an acute angle.

–

Ans:
To prove: tan²A + 1 = sec²A
Taking L.H.S., tan² A+ 1

∴ L.H.S = R.H.S

Q16: Prove that:

–

Ans:

Q17: Prove that:

–

Ans:

Q18: Prove that:

–

Ans:

Q19:

–

Ans:

Q20: Given that sinθ + cosθ = x, prove that

–

Ans: sin θ + cos θ = x …(i)
Squaring both sides in equation (i),
sin² θ + cos² θ + 2sin θ cos θ = x²
2sin θ cos θ = x² – 1 …(ii)
[∵ sin² θ + cos² θ = 1]
Also, sin² θ + cos² θ = (sin θ + cos θ)² – 2sin θ cos θ

Q21: Prove that:

–

Ans:

Previous Year Questions 2024

Q1: If sin α = √3/2, cos β = √3/2 then tan α. tan β is: (1 Mark) (CBSE 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
= √3 x 1√3

= 1

Q2: Evaluate: 5 tan 60°(sin² 60° + cos² 60°) tan 30° (3 Marks) (CBSE 2024)

–

Ans:

5 tan 60°(sin² 60° + cos² 60°) tan 30° = 5 × √31 × 1√3

= 5 × √3 × √3

= 5 × 3

= 15

Q3: Prove that: (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1 (3 Marks) (CBSE 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)
Since, cosec θ = 1/sin θ, sec θ = 1/cos θ, tan θ = sin θ/cos θ, cot θ = cos θ/sin θ
= (1/sin θ – sin θ) (1/cos θ – cos θ) (sin θ/cos θ + cos θ/sin θ)

= 1 – sin²θsin θ × 1 – cos²θcos θ × sin²θ + cos²θsin θ . cos θ

= cos²θ × sin²θsin θ × cos θ × 1sin θ . cos θ

= sin θ . cos θ1 × 1sin θ . cos θ [ : sin²θ + cos²θ = 1 ]

= 1 = R.H.S.
Hence, proved.

Previous Year Questions 2023

Q1: If 2 tan A = 3, then find the value of 4 sin A + 5 cos A6 sin A + 2 cos A is (3 Marks)(2023)

–

Ans:

Given:

2 tan A = 3 ⇒ tan A = 3/2

Using sin² A + cos² A = 1, let:

sin A = 3/√13, cos A = 2/√13

Substituting in the given expression:

4 sin A + 5 cos A6 sin A + 2 cos A

= 4 × 3/√13 + 5 × 2/√136 × 3/√13 + 2 × 2/√13

= 12/√13 + 10/√1318/√13 + 4/√13

= 22/√1322/√13

= 1

Q2: 5/8 sec²60° – tan²60° + cos²45° is equal to (1 Mark) (2023)
(a) 5/3
(b) -1/2
(c) 0
(d) -1/4

–

Ans: (c)
Sol:
58 × (2)² – (√3)² + (1√2)² = 5/8 × 4 – 3 + 12 = 0

Q3: Evaluate 2 sec²θ + 3 cosec²θ – 2 sin θ cos θ if θ = 45° (2 Marks) (CBSE 2023)

–

Ans: Since θ = 45°, sec 45° = √2, cosec 45° = √2, sin 45°= 1/√2 cos 45° = 1/√2
2sec² θ + 3 cosec² θ – 2 sin θ cos θ
= 2 (√2)² + 3 (√2)² – 2 (1√2) × (1√2)

= 2 × 2 + 3 × 2 – 2 × 12

= 4 + 6 – 1

= 9

Q4: Which of the following is true for all values of θ(0^o ≤ θ ≤ 90^o)? (1 Mark) (2023)
(a) cos²θ – sin²θ – 1
(b) cosec²θ – sec²θ- 1
(c) sec²θ – tan²θ – 1
(d) cot²θ- tan²θ = 1

–

Ans: (c)

Option (a): cos²θ – sin²θ – 1
Using the identity: cos²θ – sin²θ = cos 2θ, we get cos²θ – sin²θ – 1 = cos 2θ – 1, which is not always true. So, this option is incorrect.

Option (b): cosec²θ – sec²θ – 1
Using the identities cosec²θ = 1 + cot²θ and sec²θ = 1 + tan²θ,
we get cosec²θ – sec²θ – 1 = (1 + cot²θ) – (1 + tan²θ) – 1 = cot²θ – tan²θ – 1, which is not always zero. So, this option is incorrect.

Option (c): sec²θ – tan²θ – 1
Using the identity sec²θ = 1 + tan²θ,
we get sec²θ – tan²θ – 1 = (1 + tan²θ) – tan²θ – 1 = 0, which is always true.
So, this option is correct.

Option (d): cot²θ – tan²θ = 1
Using the identity cot²θ = 1/tan²θ, we get cot²θ – tan²θ = (1/tan²θ) – tan²θ, which is not always equal to 1. So, this option is incorrect.

Q5: If sinθ +cosθ = √3. then find the value of sinθ. cosθ. (3 Marks) (2023)

–

Ans: Given, sinθ +cosθ = √3
Squaring both sides, we get (sinθ + cosθ)² = (√3)²
⇒ sin²θ + cos²θ + 2sinθ cosθ = 3 ( ∵ sin²θ + cos²θ = 1)
⇒ 1 + 2sinθ cosθ = 3
⇒ 2sinθ cosθ = 3 – 1
⇒ 2sinθ cosθ = 2
⇒ sinθ cosθ = 1

Q6: If sin α = 1/√2 and cot β = √3, then find the value of cosec α + cosec β. (3 Marks) (2023)

–

Ans: Given, sin α = 1/√2 and cot β = √3
We know that, cosec α = 1/sinα = √2
Also, 1 + cot²β = cosec²β
⇒ cosec²β = 4
⇒ cosec β = √4 = 2
Now, cosec α + cosec β = √2 + 2

Q7: Prove that the Following Identities: Sec A (1 + Sin A) ( Sec A – tan A) = 1 (3 Marks) (2023)

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Ans: LHS = sec A(1 + sin A )( sec A – tan A)

= 1cos A (1 + sin A) 1cos A – sin Acos A

= 1cos A (1 + sin A) (1 – sin Acos A)

= 1 – sin² Acos² A = cos² Acos² A

= 1

= RHS

Hence proved..

Q8: (sec²θ – 1) (cosec² θ – 1) is equal to: (1 Mark) (CBSE 2023)
(a) –1
(b) 1
(c) 0
(d) 2

–

Ans: (b)

(sec²θ – 1) (cosec²θ – 1) = tan²θ . cot²θ

tan²θtan²θ [ ∵ sec²θ – 1 = tan²θ & cosec²θ – 1 = cot²θ ]

= 1

Q9: If sin θ – cos θ = 0, then find the value of sin⁴ θ + cos⁴ θ. (2 Marks) (CBSE 2023)

–

Ans: Given,
sin θ – cos θ = 0
sin θ = cos θ
tan θ = 1
tan θ = tan 45°
⇒ θ = 45°
Now, sin⁴ θ + cos⁴ θ = sin⁴45° + cos⁴45°

= (1√2)⁴ + (1√2)⁴

= 14 + 14 = 12

Q10: Prove that sin A – 2 sin³ A2 cos³ A – cos A = tan A (4 & 5 Marks) (CBSE 2023)

–

Ans:

LHS = sin A – 2 sin³ A2 cos³ A – cos A

= sin A (1 – 2 sin² A)cos A (2 cos² A – 1)

= sin A (1 – 2 (1 – cos² A)cos A (2 cos² A – 1)

= tan A 1 – 2 + 2 cos² A2 cos² A – 1

= tan A 2 cos² A – 12 cos² A – 1

= tan A

= RHS

Previous Year Questions 2022

Q1: Given that cos θ = √3/2, then the value of cosec²θ – sec²θcosec²θ + sec²θ3 is (2022)
(a) -1
(b) 1
(c) 1/2
(d) -1/2

–

Ans: (c)
Sol:
Given, cosθ = √3/2 = B/H

Let B = √3k and H = 2k

∴ P = √((2k)² – (√3k)²) [By Pythagoras Theorem]

⇒ k² = k

∴ cosec θ = H / p = 2k / k = 2

sec θ = H / B = 2k / √3k = 2 / √3

cosec²θ – sec²θ = (2)² – (2 / √3)²4 – 4/3

= 4 – 43 = 83

cosec²θ + sec²θ = (2)² + (2 / √3)²4 + 4/3

= 4 + 43 = 163

Q2: 1cosec θ (1 – cot θ) + 1sec θ (1 – tan θ) is equal to (2022)
(a) 0
(b) 1
(c) sinθ + cosθ
(d) sinθ – cosθ

–

Ans: (c)
Sol: We have,

1cosec θ (1 – cot θ) + 1sec θ (1 – tan θ)

= sin θcos θ / 1 – cos θsin θ + 1 – sin θ1 – cos θ

= 1cosec θ = sin θcos θ, 1sec θ

= sin² θcos² θ = sin² θ – cos² θsin θ – cos θ

= sin θ + cos θ

Q3: 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°

Q4: If sin²θ + sinθ = 1, then find the value of cos²θ + cos⁴θ is (2022)
(a) -1
(b) 1
(c) 0
(d) 2

–

Ans: (b)
Sol: Given, sin²θ + sinθ = 1 —(i)
sinθ = 1 – sin²θ
⇒ sinθ = cos²θ —(ii)
∴ cos²θ + cos⁴θ
= sinθ + sin²θ [From (ii)]
= 1 [From (i)]

Also read: Introduction: Trigonometric Ratios

Previous Year Questions 2021

Q1: If 3 sin A = 1. then find the value of sec A. (2021 C)

–

Ans: We have, 3 sin A = 1
∴ sin A = 1/3
Now by using cos²A = 1 – sin² A, we get

cos² A = 1 – 19 = 89
⇒ cos A = 2√23

∴ sec A = 1cos A = 12√2 / 3 = 3√24

Q2: Show that: 1 + cot²θ1 + tan²θ = cot²θ (2021 C)

–

Ans: We have, L.H.S.
1 + cot²θ1 + tan²θ = cosec²θsec²θ
[By using 1 + tan²θ = sec²θ and 1 + cot² θ = cosec²θ ]
⇒ 1sin²θ = cos²θsin²θ = cot²θ
Hence,
1 + cot²θ1 + tan²θ = cot²θ

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Previous Year Questions 2020

Q1: If sin θ = cos θ, then the value of tan² θ + cot² θ is (2020)
(a) 2
(b) 4
(c) 1
(d) 10/3

–

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]
∴ tan²θ + cot²θ = 1² + 1² = 2
Hence, A option is correct.

Q2: Given 15 cot A = 8, then find the values of sin A and sec A. (2020)

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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
AC²= AB² + BC²
⇒ (8k)² + (15)² = 64k² + 225k² = 289k² = (17k)²

⇒ AC = √((17k)²) = 17k

∴ sin A = BCAC = 15k17k = 1517

and cos A = ABAC = 8k17k = 817

So, sec A = 1cos A = 178

Q3: Write the value of sin² 30° + cos² 60°. (2020)

–

Ans: We have, sin² 30° + cos² 60°

Q4: The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ − b cos θ) is (2020)
(a) a²+ b²
(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

AB = √(x₂ – x₁)² + (y₂ – y₁)²

So,

AB = √(a cos θ + b sin θ – 0)² + (0 – a sin θ + b cos θ)²

= √ a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ)

But according to the trigonometric identity,

sin²θ + cos²θ = 1

Therefore,

AB = √ a² + b²

Q5: 5 tan²θ – 5 sec²θ = ____________. (2020)

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Ans: We have 5(tan²θ – sec²θ)
= 5(-1) = – 5 [By using 1 + tan²θ = sec² θ ⇒ tan²θ – sec²θ = – 1]

Q6: If sinθ + cosθ = √3. then prove that tan θ + cot θ = 1 (2020)

–

Ans: sin θ + cos θ =√3
= (sinθ + cosθ)²= (√3)²
= sin² θ + cos² θ + 2sin θ cos θ = 3 (Since,sin²θ + cos²θ = 1)
= 1 + 2sin θ cos θ = 3
⇒ 2sin θ cos θ = 2
⇒ sin θ cos θ = 1
⇒ sin θ cos θ = sin²θ + cos²θ
⇒ 1 = sin²θ + cos²θsin θ cos θ
⇒ tan θ + cot θ = 1

Q7: If x = a sinθ and y = b cosθ, write the value of (b²x² + a²y²). (CBSE 2020)

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Ans: Given, x = a sin θ and y = b cos θ
Putting the values of x and y in (b²x² + a²y²)
We get,
= b²(a sin θ)²+ a²_{(b cos θ)}²
= a²b² [sin² θ + cos² θ] [Also, sin²θ + cos²θ = 1]
= a²b² [1]
= a²b²

Q8: Prove that: 2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1 = 0. (CBSE 2020)

–

Ans: We know that,
sin² θ + cos² θ = 1
So, (sin² θ + cos² θ) ² = 1²
⇒ sin⁴ θ + cos⁴θ + 2sin² θ cos² θ = 1
i.e., sin⁴ θ + cos⁴ θ = 1 – 2 sin² θ cos² θ …(i)
Also, (sin² θ + cos² θ) ³ = 1³
⇒ sin⁶ θ + cos⁶ θ + 3 sin² θ cos² θ (sin² θ + cos² θ) = 1
⇒ sin⁶ θ+ cos⁶ θ+ 3sin2 θ cos² θ (1) = 1
i.e., sin⁶ θ + cos⁶ θ = 1 – 3 sin² θ cos² θ …(ii)
Now,
LHS = 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1
= 2(1 – 3 sin² θ cos² θ) – 3(1 – 2 sin² θ cos² θ) + 1
= 2 – 3 + 1
= 0
Hence, proved.

Q9: Prove that: (sin⁴ θ – cos⁴ θ + 1) cosec² θ = 2. [CBSE 2020].

–

Ans: L.H.S. = (sin⁴ θ – cos⁴ θ + 1) cosec² θ
= [(sin² θ + cos² θ) (sin² θ – cos² θ) + 1] cosec² θ
[(1) (sin² θ – cos² θ) + 1] cosec²θ as [ sin² θ + cos² θ = 1]
= [sin² θ + (1 – cos² θ)] cosec² θ
= (sin²θ + sin²θ) cosec²θ
= (2 sin² θ) cosec² θ
=

= 2 × 1
= 2 = R.H.S.
Hence, proved.

Previous Year Questions 2019

Q1: If sin x + cos y = 1, x = 30° and y is an acute angle, find the value of y. (2019)

–

Ans: 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°.

Q2: If cosec² 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
Sin² α + Cos² α = 1
(a + b)(a – b) = a² – b²
Calculation:
cosec2 θ (cos θ – 1)(1 + cos θ) = k
⇒ cosec2 θ (1 – cos θ)(1 + cos θ) = – k
⇒ cosec2 θ (1 – cos² θ) = -k [Also, sin² θ + cos² θ = 1]
⇒ cosec2 θ × sin² θ = -k
⇒ 1 = -k
⇒ k = -1
∴ The value of k is (-1).

Q3: The value of ( 1 + cot A − cosec A ) ( 1 + tan A + sec A ) is

–

Ans:

(1 + cos Asin A – 1sin A ) (1 + sin Acos A + 1cos A )

= sin A + cos A – 1sin A × cos A + sin A + 1cos A

= (sin A + cos A)² – 1sin A . cos A

= sin²A + cos²A + 2 sin A . cos A – 1sin A . cos A

= sin²A + cos²A – 1 + 2 sin A . cos Asin A . cos A

= 2

Also read: Introduction: Trigonometric Ratios

Previous Year Questions 2013

Q1: If sec θ + tan θ + 1 = 0, then sec θ – tan θ is:
(a) –1
(b) 1
(c) 0
(d) 2 (CBSE 2013)

–

Ans: (a)

sec θ + tan θ + 1 = 0

⇒ sec θ + tan θ = -1

Multiplying and dividing LHS by sec θ – tan θ, we get

⇒ (sec θ + tan θ) × sec θ – tan θsec θ – tan θ = -1

⇒ sec² θ – tan² θsec θ – tan θ = -1

⇒ 1 + tan² θ – tan² θsec θ – tan θ = -1 (∵ sec² θ = 1 + tan² θ)

⇒ 1sec θ – tan θ = -1

⇒ sec θ – tan θ = -1

Hence, the correct option is (a).

Previous Year Questions: Introduction to Trigonometry

Previous Year Questions 2025

Q1: If tan A+ cot A= 6, then find the value of tan²A + cot² A – 4.

–

Q2: If tan 3θ = √3, then θ/2 equals
(a) 60°
(b) 30°
(c) 20°
(d) 10°

–

Ans: (d)
We have, tan3θ = √3
⇒ tan3θ = tan 60°

Q3: If sin 4θ = √3/2, then θ/3 equals:
(a) 60°
(b) 20°
(c) 15°
(d) 5°

–

Ans: (d)
Given, sin 4θ = √3/2 = sin 60°.
⇒ 4θ = 60°
⇒ θ = 15°.

Q4: If α + β = 90° and α = 2β, then cos²α + sin²β is equal to:
(a) 0
(b) 1/2
(c) 1
(d) 2

–

Ans: (b)
α + β = 90° and α = 2β …(i)
∴ 2β + β = 90° ⇒ 3β = 90° ⇒ β = 30°
∴ α = 2 × 30° = 60° [From (i)]
∴ cos²α + sin²β = cos²60° + sin²30°

Q5: then x : y =
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 4: 1

–

Ans: (c)

Q6: If 4k = tan²60° – 2cosec² 30° – 2tan² 30°, then find the value of k.

–

Ans: Given, 4k = tan²60° – 2 cosec²30° – 2 tan²30°

Q7: If x cos60° + ycos0° + sin30° – cot45° = 5, then find the value of x + 2y.

–

Ans: We have, x cos60° + y cos0° + sin30° – cot45° = 5

Q8:

–

Ans:

Q9:
(a) cot θ
(b)
(c)
(d) tan θ

–

Ans: (a)
We have,
[∵ 1 – cos²θ = sin²θ]

Q10: The value of (tan A cosec A)² – (sin A sec A)² is:
(a) 0
(b) 1
(c) -1
(d) 2

–

Ans: (b)
We have,
(tan A · cosec A)² – (sin A . sec A)²
Using identities:
tan A = sin A / cos A,
cosec A = 1 / sin A,
sec A = 1 / cos A.
sec² A – tan² A = 1

Q11: (cotθ + tanθ) equals:
(a) cosecθ secθ
(b) sinθ secθ
(c) cosθ tanθ
(d) sinθ cosθ

–

Ans: (a)
We have, cotθ + tanθ
cot θ = cos θ / sin θ,
tan θ = sin θ / cos θ.

Q12: The value of
(a) 1
(b) 0
(c) -1
(d) 2

–

Ans: (c)

Q13: In a right triangle ABC, right-angled at A, if sin B = 1/4 then the value of sec B is
(a) 4
(b) √15/4
(c) √15
(d) 4/√15

–

Ans: (d)
Given, sin B = 1/4

Q14: If a secθ + b tan θ = m and b sec θ + a tan θ = n, prove that a² + n² = b² + m²

–

Q15: Use the identity: sin²A + cos²A = 1 to prove that tan²A + 1 = sec²A. Hence, find the value of tan A, where sec A = 5/3, where A is an acute angle.

–

Ans:
To prove: tan²A + 1 = sec²A
Taking L.H.S., tan² A+ 1

∴ L.H.S = R.H.S

Q16: Prove that:

–

Ans:

Q17: Prove that:

–

Ans:

Q18: Prove that:

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Ans:

Q19:

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Ans:

Q20: Given that sinθ + cosθ = x, prove that

–

Q21: Prove that:

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Ans:

Previous Year Questions 2024

Q1: If sin α = √3/2, cos β = √3/2 then tan α. tan β is: (1 Mark) (CBSE 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
= √3 x 1√3

= 1

Q2: Evaluate: 5 tan 60°(sin² 60° + cos² 60°) tan 30° (3 Marks) (CBSE 2024)

–

Ans:

5 tan 60°(sin² 60° + cos² 60°) tan 30° = 5 × √31 × 1√3

= 5 × √3 × √3

= 5 × 3

= 15

Q3: Prove that: (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1 (3 Marks) (CBSE 2024)

–

= 1 – sin²θsin θ × 1 – cos²θcos θ × sin²θ + cos²θsin θ . cos θ

= cos²θ × sin²θsin θ × cos θ × 1sin θ . cos θ

= sin θ . cos θ1 × 1sin θ . cos θ [ : sin²θ + cos²θ = 1 ]

= 1 = R.H.S.
Hence, proved.

Previous Year Questions 2023

Q1: If 2 tan A = 3, then find the value of 4 sin A + 5 cos A6 sin A + 2 cos A is (3 Marks)(2023)

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Ans:

Given:

2 tan A = 3 ⇒ tan A = 3/2

Using sin² A + cos² A = 1, let:

sin A = 3/√13, cos A = 2/√13

Substituting in the given expression:

4 sin A + 5 cos A6 sin A + 2 cos A

= 4 × 3/√13 + 5 × 2/√136 × 3/√13 + 2 × 2/√13

= 12/√13 + 10/√1318/√13 + 4/√13

= 22/√1322/√13

= 1

Q2: 5/8 sec²60° – tan²60° + cos²45° is equal to (1 Mark) (2023)
(a) 5/3
(b) -1/2
(c) 0
(d) -1/4

–

Ans: (c)
Sol:
58 × (2)² – (√3)² + (1√2)² = 5/8 × 4 – 3 + 12 = 0

Q3: Evaluate 2 sec²θ + 3 cosec²θ – 2 sin θ cos θ if θ = 45° (2 Marks) (CBSE 2023)

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Ans: Since θ = 45°, sec 45° = √2, cosec 45° = √2, sin 45°= 1/√2 cos 45° = 1/√2
2sec² θ + 3 cosec² θ – 2 sin θ cos θ
= 2 (√2)² + 3 (√2)² – 2 (1√2) × (1√2)

= 2 × 2 + 3 × 2 – 2 × 12

= 4 + 6 – 1

= 9

–

Ans: (c)

Option (a): cos²θ – sin²θ – 1
Using the identity: cos²θ – sin²θ = cos 2θ, we get cos²θ – sin²θ – 1 = cos 2θ – 1, which is not always true. So, this option is incorrect.

Option (d): cot²θ – tan²θ = 1
Using the identity cot²θ = 1/tan²θ, we get cot²θ – tan²θ = (1/tan²θ) – tan²θ, which is not always equal to 1. So, this option is incorrect.

Q5: If sinθ +cosθ = √3. then find the value of sinθ. cosθ. (3 Marks) (2023)

–

Q6: If sin α = 1/√2 and cot β = √3, then find the value of cosec α + cosec β. (3 Marks) (2023)

–

Ans: Given, sin α = 1/√2 and cot β = √3
We know that, cosec α = 1/sinα = √2
Also, 1 + cot²β = cosec²β
⇒ cosec²β = 4
⇒ cosec β = √4 = 2
Now, cosec α + cosec β = √2 + 2

Q7: Prove that the Following Identities: Sec A (1 + Sin A) ( Sec A – tan A) = 1 (3 Marks) (2023)

–

Ans: LHS = sec A(1 + sin A )( sec A – tan A)

= 1cos A (1 + sin A) 1cos A – sin Acos A

= 1cos A (1 + sin A) (1 – sin Acos A)

= 1 – sin² Acos² A = cos² Acos² A

= 1

= RHS

Hence proved..

Q8: (sec²θ – 1) (cosec² θ – 1) is equal to: (1 Mark) (CBSE 2023)
(a) –1
(b) 1
(c) 0
(d) 2

–

Ans: (b)

(sec²θ – 1) (cosec²θ – 1) = tan²θ . cot²θ

tan²θtan²θ [ ∵ sec²θ – 1 = tan²θ & cosec²θ – 1 = cot²θ ]

= 1

Q9: If sin θ – cos θ = 0, then find the value of sin⁴ θ + cos⁴ θ. (2 Marks) (CBSE 2023)

–

Ans: Given,
sin θ – cos θ = 0
sin θ = cos θ
tan θ = 1
tan θ = tan 45°
⇒ θ = 45°
Now, sin⁴ θ + cos⁴ θ = sin⁴45° + cos⁴45°

= (1√2)⁴ + (1√2)⁴

= 14 + 14 = 12

Q10: Prove that sin A – 2 sin³ A2 cos³ A – cos A = tan A (4 & 5 Marks) (CBSE 2023)

–

Ans:

LHS = sin A – 2 sin³ A2 cos³ A – cos A

= sin A (1 – 2 sin² A)cos A (2 cos² A – 1)

= sin A (1 – 2 (1 – cos² A)cos A (2 cos² A – 1)

= tan A 1 – 2 + 2 cos² A2 cos² A – 1

= tan A 2 cos² A – 12 cos² A – 1

= tan A

= RHS

Previous Year Questions 2022

Q1: Given that cos θ = √3/2, then the value of cosec²θ – sec²θcosec²θ + sec²θ3 is (2022)
(a) -1
(b) 1
(c) 1/2
(d) -1/2

–

Ans: (c)
Sol:
Given, cosθ = √3/2 = B/H

Let B = √3k and H = 2k

∴ P = √((2k)² – (√3k)²) [By Pythagoras Theorem]

⇒ k² = k

∴ cosec θ = H / p = 2k / k = 2

sec θ = H / B = 2k / √3k = 2 / √3

cosec²θ – sec²θ = (2)² – (2 / √3)²4 – 4/3

= 4 – 43 = 83

cosec²θ + sec²θ = (2)² + (2 / √3)²4 + 4/3

= 4 + 43 = 163

Q2: 1cosec θ (1 – cot θ) + 1sec θ (1 – tan θ) is equal to (2022)
(a) 0
(b) 1
(c) sinθ + cosθ
(d) sinθ – cosθ

–

Ans: (c)
Sol: We have,

1cosec θ (1 – cot θ) + 1sec θ (1 – tan θ)

= sin θcos θ / 1 – cos θsin θ + 1 – sin θ1 – cos θ

= 1cosec θ = sin θcos θ, 1sec θ

= sin² θcos² θ = sin² θ – cos² θsin θ – cos θ

= sin θ + cos θ

Q3: 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°

Q4: If sin²θ + sinθ = 1, then find the value of cos²θ + cos⁴θ is (2022)
(a) -1
(b) 1
(c) 0
(d) 2

–

Ans: (b)
Sol: Given, sin²θ + sinθ = 1 —(i)
sinθ = 1 – sin²θ
⇒ sinθ = cos²θ —(ii)
∴ cos²θ + cos⁴θ
= sinθ + sin²θ [From (ii)]
= 1 [From (i)]

Also read: Introduction: Trigonometric Ratios

Previous Year Questions 2021

Q1: If 3 sin A = 1. then find the value of sec A. (2021 C)

–

Ans: We have, 3 sin A = 1
∴ sin A = 1/3
Now by using cos²A = 1 – sin² A, we get

cos² A = 1 – 19 = 89
⇒ cos A = 2√23

∴ sec A = 1cos A = 12√2 / 3 = 3√24

Q2: Show that: 1 + cot²θ1 + tan²θ = cot²θ (2021 C)

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Previous Year Questions 2020

Q1: If sin θ = cos θ, then the value of tan² θ + cot² θ is (2020)
(a) 2
(b) 4
(c) 1
(d) 10/3

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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]
∴ tan²θ + cot²θ = 1² + 1² = 2
Hence, A option is correct.

Q2: Given 15 cot A = 8, then find the values of sin A and sec A. (2020)

–

⇒ AC = √((17k)²) = 17k

∴ sin A = BCAC = 15k17k = 1517

and cos A = ABAC = 8k17k = 817

So, sec A = 1cos A = 178

Q3: Write the value of sin² 30° + cos² 60°. (2020)

–

Ans: We have, sin² 30° + cos² 60°

Q4: The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ − b cos θ) is (2020)
(a) a²+ b²
(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

AB = √(x₂ – x₁)² + (y₂ – y₁)²

So,

AB = √(a cos θ + b sin θ – 0)² + (0 – a sin θ + b cos θ)²

= √ a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ)

But according to the trigonometric identity,

sin²θ + cos²θ = 1

Therefore,

AB = √ a² + b²

Q5: 5 tan²θ – 5 sec²θ = ____________. (2020)

–

Ans: We have 5(tan²θ – sec²θ)
= 5(-1) = – 5 [By using 1 + tan²θ = sec² θ ⇒ tan²θ – sec²θ = – 1]

Q6: If sinθ + cosθ = √3. then prove that tan θ + cot θ = 1 (2020)

–

Q7: If x = a sinθ and y = b cosθ, write the value of (b²x² + a²y²). (CBSE 2020)

–

Q8: Prove that: 2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1 = 0. (CBSE 2020)

–

Q9: Prove that: (sin⁴ θ – cos⁴ θ + 1) cosec² θ = 2. [CBSE 2020].

–

= 2 × 1
= 2 = R.H.S.
Hence, proved.

Previous Year Questions 2019

Q1: If sin x + cos y = 1, x = 30° and y is an acute angle, find the value of y. (2019)

–

Ans: 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°.

Q2: If cosec² θ (cos θ – 1)(1 + cos θ) = k, then what is the value of k? (2019)

–

Q3: The value of ( 1 + cot A − cosec A ) ( 1 + tan A + sec A ) is

–

Ans:

(1 + cos Asin A – 1sin A ) (1 + sin Acos A + 1cos A )

= sin A + cos A – 1sin A × cos A + sin A + 1cos A

= (sin A + cos A)² – 1sin A . cos A

= sin²A + cos²A + 2 sin A . cos A – 1sin A . cos A

= sin²A + cos²A – 1 + 2 sin A . cos Asin A . cos A

= 2

Also read: Introduction: Trigonometric Ratios

Previous Year Questions 2013

Q1: If sec θ + tan θ + 1 = 0, then sec θ – tan θ is:
(a) –1
(b) 1
(c) 0
(d) 2 (CBSE 2013)

–

Ans: (a)

sec θ + tan θ + 1 = 0

⇒ sec θ + tan θ = -1

Multiplying and dividing LHS by sec θ – tan θ, we get

⇒ (sec θ + tan θ) × sec θ – tan θsec θ – tan θ = -1

⇒ sec² θ – tan² θsec θ – tan θ = -1

⇒ 1 + tan² θ – tan² θsec θ – tan θ = -1 (∵ sec² θ = 1 + tan² θ)

⇒ 1sec θ – tan θ = -1

⇒ sec θ – tan θ = -1

Hence, the correct option is (a).