Trigonometry MCQ Quiz - Objective Question with Answer for Trigonometry - Download Free PDF
Last updated on Jun 27, 2025
Latest Trigonometry MCQ Objective Questions
Trigonometry Question 1:
The number of solutions of the equation
Answer (Detailed Solution Below)
Trigonometry Question 1 Detailed Solution
Concept:
- Range of Sine Function: The output of the sine function is always between −1 and 1, i.e., sin(θ) ∈ [−1, 1] for all real θ.
- Quadratic Function: A quadratic function of the form ax2 + bx + c represents a parabola. If a > 0, the parabola opens upward, and its minimum value occurs at x = −b / 2a.
- Key Idea: To find how many solutions exist for sin(expression) = quadratic, we determine how many values of x make the quadratic expression lie within [−1, 1].
Calculation:
Given,
Let f(x) = x2 − 4x + 6
Minimum of f(x) occurs at:
x = 4 / 2 = 2
⇒ f(2) = (2)2 − 4×2 + 6 = 4 − 8 + 6 = 2
Since the parabola opens upward, the range of f(x) is [2, ∞)
But, sin(θ) ∈ [−1, 1]
⇒ The equation has solutions only if x2 − 4x + 6 ∈ [−1, 1]
But f(x) ≥ 2 for all x, and 2 > 1
⇒ No value of x satisfies f(x) ∈ [−1, 1]
∴ Number of real solutions is zero.
Trigonometry Question 2:
Comprehension:
The value of (1 + tan x) (1 + tan y) is
Answer (Detailed Solution Below)
Trigonometry Question 2 Detailed Solution
Concept:
Solution:
Given:
If x, y, and z are the angles of a triangle and z = 135°
⇒ x + y + z = 180o
⇒ x + y = 180o - 135o
⇒ x + y = 45o
⇒ tan (x + y) = tan (45o)
⇒
⇒ tan x + tan y = 1 - tan x tan y
Adding 1 both side,
⇒ 1 + tan x + tan y = 1 - tan x tan y + 1
⇒ 1 + tan x + tan y + tan x tan y = 2
⇒ 1 + tan x + tan y(1 + tan x) = 2
⇒ (1 + tan x) (1+ tan y) = 2
∴ The value of (1 + tan x) (1 + tan y) is 2
Trigonometry Question 3:
Comprehension:
The value of sin z + cos z is
Answer (Detailed Solution Below)
Trigonometry Question 3 Detailed Solution
Calculation:
We can rewrite the terms as:
Using the standard trigonometric identities
Now, substitute the values:
Thus, we get:
Hence, the correct answer is Option 1.
Trigonometry Question 4:
Comprehension:
What is the maximum area of the triangle?
Answer (Detailed Solution Below)
Trigonometry Question 4 Detailed Solution
Calculation:
Given,
Let
Then,
The area of the triangle is,
To maximize, differentiate w.r.t.
At
∴ The maximum area is
Hence, the correct answer is Option 1.
Trigonometry Question 5:
Comprehension:
What is ∠A equal to if the area of the triangle is maximum?
Answer (Detailed Solution Below)
Trigonometry Question 5 Detailed Solution
Calculation:
Given,
Let
Then,
The area of the triangle is,
To maximize, set up
Hence
Therefore,
∴
Hence, the correct answer is Option 3.
Top Trigonometry MCQ Objective Questions
Find the value of cos 47° sec 133° + sin 44° cosec 136°.
Answer (Detailed Solution Below)
Trigonometry Question 6 Detailed Solution
Download Solution PDFFormula used:
sec (180° - θ) = - sec θ
cosec (180° - θ) = cosec θ
cos θ × sec θ = 1 ; sin θ × cosec θ = 1
Calculation:
cos 47° sec 133° + sin 44° cosec 136°
⇒ cos 47° × sec (180° - 47) + sin 44° cosec (180° - 44°)
⇒ cos 47° × (- sec 47°) + sin 44° × (cosec 44°)
⇒ -1 + 1 = 0
∴ The correct answer is 0.
Answer (Detailed Solution Below)
Trigonometry Question 7 Detailed Solution
Download Solution PDFGiven:
Concept used:
Calculation:
⇒
⇒
⇒
⇒
⇒
⇒
⇒
∴ The required answer is
The value of tan2θ + cot2θ - sec2θ cosec2θ is:
Answer (Detailed Solution Below)
Trigonometry Question 8 Detailed Solution
Download Solution PDFGiven:
tan2θ + cot2θ - sec2θ cosec2θ
Concept used:
1. tanα = sinα/cosα
2. cotα = 1/tanα
3. secα = 1/cosα
4. cosecα = 1/sinα
5. (a + b)2 - 2ab = a2 + b2
6. sin2α + cos2α = 1
Calculation:
tan2θ + cot2θ - sec2θ cosec2θ
⇒
⇒
⇒
⇒
⇒
⇒ -2
∴ The required answer is -2.
Shortcut Trick
Use value putting method to solve this question,
Use θ = 45°
tan2θ + cot2θ - sec2θ cosec2θ
⇒ 12 + 12 - (√2)2(√2)2
⇒ 1 + 1 - 4
⇒ 2 - 4 = - 2
∴ The correct answer to this question is -2.
If sec θ + tan θ = 5, then find the value of tan θ.
Answer (Detailed Solution Below)
Trigonometry Question 9 Detailed Solution
Download Solution PDFGiven:
sec θ + tan θ = 5
Concept used:
If sec θ + tan θ = y
then sec θ - tan θ = 1/y
Calculation:
sec θ + tan θ = 5 ----- (1)
then,
sec θ - tan θ = 1/5 ------- (2)
Subtracting the eq. (1) and (2)
⇒ (sec θ + tan θ) - (sec θ - tan θ) = (5 - 1/5)
⇒ sec θ + tan θ - sec θ + tan θ = 24/5
⇒ 2 × tan θ = 24/5
⇒ tan θ = 12/5
∴ The correct answer is 12/5.
Evaluate the following:
cos(36° + A).cos(36° - A) + cos(54° + A).cos(54° - A)
Answer (Detailed Solution Below)
Trigonometry Question 10 Detailed Solution
Download Solution PDFGiven:
cos (36° - A) cos (36° + A) + cos (54° - A) cos (54° + A)
Formula used:
cos (a - b) = cos a cos b + sin a sin b.
sin (90 - a) = cos a
Calculation:
⇒ sin[90 – (36 – A)]sin[90 – (36 + A)] + cos (54° – A) cos (54° + A)
⇒ sin(54º + A)sin(54º – A) + cos (54° – A)cos (54° + A)
⇒ Using the identity cos(A – B),
⇒ cos(54 + A – 54 + A) = cos(2A)
Therefore, the value of cos (36° - A) cos (36° + A) + cos (54° - A) cos (54° + A) is cos(2A).
Find the value of cos 2A cos 2B + sin2(A - B) - sin2(A + B)
Answer (Detailed Solution Below)
Trigonometry Question 11 Detailed Solution
Download Solution PDFGiven:
cos 2A cos 2B + sin2(A - B) - sin2(A + B)
Concept used:
cos (a + b) = cos a cos b - sin a sin b
sin2a - sin2b = sin(a + b) sin(a - b)
Calculation:
cos 2A cos 2B + sin2(A - B) - sin2(A + B)
⇒ cos 2A cos 2B - [sin2(A + B) - sin2(A - B)]
{sin2a - sin2b = sin(a + b) sin(a - b)}
⇒ cos 2A cos 2B - [sin(A + B + A - B) sin(A + B - A + B)]
⇒ cos 2A cos 2B - [sin(A + A) sin(B + B)]
⇒ cos 2A cos 2B - sin 2A sin 2B
⇒ cos (2A + 2B)
∴ The required answer is cos (2A + 2B).
Find the value of sin (1920°)
Answer (Detailed Solution Below)
Trigonometry Question 12 Detailed Solution
Download Solution PDFConcept:
sin (2nπ ± θ) = ± sin θ
sin (90 + θ) = cos θ
Calculation:
Given: sin (1920°)
⇒ sin (1920°) = sin(360° × 5° + 120°) = sin (120°)
⇒ sin (120°) = sin (90° + 30°) = cos 30° = √3 / 2If {(3 sin θ – cos θ) / (cos θ + sin θ)} = 1, then the value of cot θ is:
Answer (Detailed Solution Below)
Trigonometry Question 13 Detailed Solution
Download Solution PDFGiven:
{(3Sinθ – Cosθ)/(Cosθ + Sinθ)} = 1
Calculation:
We have a trigonometric equation
{(3Sinθ – Cosθ)/(Cosθ + Sinθ)} = 1
On dividing numerator and denominator by sin θ, we get
⇒ [{(3sinθ – cosθ)/Sinθ}/{(cosθ + sinθ)/sinθ}] = 1
⇒ {(3 – cotθ)/(cotθ + 1)} = 1
⇒ 3 – cotθ = 1 + cotθ
⇒ 2cotθ = 2
cotθ = 1
The value is 1.
Answer (Detailed Solution Below)
Trigonometry Question 14 Detailed Solution
Download Solution PDFGiven:
sin 2x + 2 sin 4x + sin 6x
Formula used:
sin C + sin D = 2 × sin (C + D)/2 × cos (C - D)/2
cos 2θ = (2 × cos 2 θ - 1)
Calculation:
sin 6x + sin 2x + 2 sin 4x
⇒ 2 × sin (6x + 2x)/2 × cos (6x - 2x)/2 + 2 sin 4x
⇒ 2 × sin 4x × cos 2x + 2 sin 4x
⇒ 2 × sin 4x (cos 2x + 1)
⇒ 2 × sin 4x {(2 × cos2x - 1) + 1) }
⇒ (2 × sin 4x) × (2 × cos2 x)
⇒ 4 cos2 x sin 4x
∴ The correct answer is 4 cos2 x sin 4x.
If a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 - q2 is equal to ______.
Answer (Detailed Solution Below)
Trigonometry Question 15 Detailed Solution
Download Solution PDFGiven:
a cot θ + b cosec θ = p
b cot θ + a cosec θ = q
Formula used:
Cosec2 θ - cot2 θ = 1
Calculation:
a cot θ + b cosec θ = p
Squaring both sides
(a cot θ + b cosec θ)2 = (p)2
a2 cot2 θ + b2 cosec2 θ + 2 × ab cot θ × cosec θ = p2 ----- (1)
b cot θ + a cosec θ = q
Squaring both sides
(b cot θ + a cosec θ)2 = (q)2
b2 cot2 θ + a2 cosec2 θ + 2 × ab cot θ × cosec θ = q2 ----- (2)
Substracting the eq.(1) and (2)
⇒ (p2 - q2) = a2 cot2 θ + b2 cosec2 θ + 2 × ab × cot θ × cosec θ - (b2 cot2 θ + a2 cosec2 θ + 2 × ab × cot θ × cosec θ)
⇒ a2 cot2 θ - a2 cosec2 θ + b2 cosec2 θ - b2 cot2 θ
⇒ a2 (cot2 θ - cosec2 θ) + b2 (cosec2 θ - cot2 θ)
⇒ b2 - a2