Trigonometry MCQ Quiz - Objective Question with Answer for Trigonometry - Download Free PDF

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,

\(\sin (\frac{\pi x }{3\sqrt{2}}) = x^2-4x+6 \)   

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.

Concept:

\(tan (A+B)= \frac{tanA+tanB}{1-tanA .tanB} \)

Solution:

Given:

If x, y, and z are the angles of a triangle and z = 135°

⇒ x + y + z = 180^o

⇒ x + y = 180^o - 135^o

⇒ x + y = 45o

⇒ tan (x + y) = tan (45o)

⇒ \(\frac{tanx+tany}{1-tanx .tany} = 1\)

⇒ 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

Calculation:

\( \sin z + \cos z = \sin \frac{3\pi}{4} + \cos \frac{3\pi}{4} \)

We can rewrite the terms as:

\( \sin \frac{3\pi}{4} = \sin \left( \pi - \frac{\pi}{4} \right) \) and \( \cos \frac{3\pi}{4} = \cos \left( \pi - \frac{\pi}{4} \right) \)

Using the standard trigonometric identities \(\sin (\pi - \theta) = \sin \theta \) and \(\cos (\pi - \theta) = -\cos \theta \), we get:

\( \sin \frac{3\pi}{4} = \sin \frac{\pi}{4} \) and \( \cos \frac{3\pi}{4} = -\cos \frac{\pi}{4} \)

Now, substitute the values:

\( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \) and \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)

Thus, we get:

\( \sin z + \cos z = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0 \)

Hence, the correct answer is Option 1.

Calculation:

Given,

\(AB + AC = 3\)

Let \(AB = x\) and \(AC = 3 - x \).

Then,

\(BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x} \)

The area of the triangle is,

\(A = \tfrac12\,x\,BC = \tfrac12\,x\,\sqrt{9 - 6x}\)

To maximize, differentiate w.r.t. \(x\) and set to zero:

\(\displaystyle \frac{d}{dx}\bigl(x\sqrt{9-6x}\bigr) = \sqrt{9-6x} \;-\;\frac{6x}{2\sqrt{9-6x}} = 0 \;\Longrightarrow\; x = 1 \)

At \(x = 1\), we get \(BC = \sqrt{9 - 6} = \sqrt{3}\), so

\(A_{\max} = \tfrac12 \times 1 \times \sqrt{3} = \frac{\sqrt{3}}{2}\)

∴ The maximum area is \(\frac{\sqrt{3}}{2}\) square unit.

Hence, the correct answer is Option  1.

Calculation:

Given,

\(AB + AC = 3\)

Let \(AB = x\) and \(AC = 3 - x\).

Then,

\(BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x} \)

The area of the triangle is,

\(A = \tfrac12\,x\,\sqrt{9 - 6x} \)

To maximize, set up

\(A^2 = \tfrac14\,x^2\,(9 - 6x)\)

\(\displaystyle \frac{d(A^2)}{dx} = \tfrac14\bigl(2x(9 - 6x) + x^2(-6)\bigr) = \frac{18x(1 - x)}{4} = 0 \)

Hence \(x = 1\) (discarding x=0, so

\(BC = \sqrt{9 - 6}= \sqrt{3},\quad AC = 3 - 1 = 2\)

Therefore,

\(\sin A = \frac{BC}{AC} = \frac{\sqrt{3}}{2} \implies A = \frac{\pi}{3}\)

∴ \(\angle A = \frac{\pi}{3}\).

Hence, the correct answer is Option 3.

Formula 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.

Given:

\(\frac{\cos 45^{\circ }}{\sec 30^{\circ}+ cosec30^{\circ}}\)

Concept used:

Calculation:

\(\frac{\cos 45^{\circ }}{\sec 30^{\circ}+ cosec30^{\circ}}\)

⇒ \(\frac {\frac {1}{\sqrt2}} {\frac {2}{\sqrt3}+ \frac {2}{1}}\)

⇒ \(\frac {\frac {1}{\sqrt2}} {2(\frac {\sqrt3 + 1}{\sqrt3})}\)

⇒ \(\frac {\sqrt3} {2{\sqrt2}({\sqrt3 + 1})}\)

⇒ \(\frac {\sqrt3({\sqrt3 - 1})} {2{\sqrt2}({\sqrt3 + 1})({\sqrt3 - 1})}\)

⇒ \(\frac {\sqrt3({\sqrt3 - 1})} {2{\sqrt2}({3 - 1)}}\)

⇒ \(\frac {({3 - \sqrt3})} {4{\sqrt2}}\)

⇒ \(\frac {({3\sqrt2 - \sqrt6})} {8}\)

∴ The required answer is \(\frac {({3\sqrt2 - \sqrt6})} {8}\).

Given:

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

⇒ \(\frac {sin^2θ}{cos^2θ} + \frac {cos^2θ}{sin^2θ} - \frac {1}{sin^2θ \times cos^2θ}\)

⇒ \(\frac {sin^4θ + cos^4θ - 1}{sin^2θ \times cos^2θ}\)

⇒ \(\frac {(sin^2θ + cos^2θ)^2 - 2sin^2θ cos^2θ - 1}{sin^2θ \times cos^2θ}\)

⇒ \(\frac {(1)^2 - 2sin^2θ cos^2θ - 1}{sin^2θ \times cos^2θ}\)

⇒ \(\frac {-2sin^2θ cos^2θ}{sin^2θ \times cos^2θ}\)

⇒ -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.

Given:

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.

Given:

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).

Given:

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(54^º + A)sin(54^º – 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).

Concept:

sin (2nπ ± θ) = ±  sin θ

sin (90 + θ) = cos θ

Calculation:

Given: sin (1920°)

⇒ sin (1920°) = sin(360° × 5° + 120°) = sin (120°)

Given:

{(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.

Given:

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.

Given:

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