Trigonometric Values MCQ Quiz - Objective Question with Answer for Trigonometric Values - Download Free PDF

Given:

Let PQR be a right-angled triangle, right-angled at R.

PQ = 29 cm, QR = 21 cm, and ∠Q = θ.

Formula used:

cos² θ - sin² θ = cos 2θ

Calculations:

In Triangle PQR using the Pythagorean theorem:

⇒ PR² = PQ² - QR²

⇒ PR² = 29² - 21²

⇒ PR² = 841 - 441

⇒ PR = 20 cm

cos θ = QR / PQ = 21 / 29

sin θ = PR / PQ = 20 / 29

Next, we use the identity for cos² θ - sin² θ:

⇒ cos² θ - sin² θ = (cos θ + sin θ) × (cos θ - sin θ)

⇒ cos² θ - sin² θ = (21/29)² - (20/29)²

⇒ cos² θ - sin² θ = (441/841) - (400/841)

⇒ cos² θ - sin² θ = (441 - 400) / 841 = 41 / 841

∴ The correct answer is option (1).

Given:

Let ABC be a right-angled triangle with a right angle at B.

tan A = √3,

Formula used:

sin (A + C) = sin A cos C + cos A sin C

cos (A + C) = cos A cos C - sin A sin C

Calculation:

Given tan A = √3

⇒ A = 60° (since tan 60° = √3)

In ΔABC, since B = 90°, A + C = 90°

⇒ C = 30°

Now, sin A cos C + cos A sin C = sin (A + C)

⇒ sin 90° = 1

cos A cos C - sin A sin C = cos (A + C)

⇒ cos 90° = 0

∴ The correct answer is option (3).

Given:

We need to find the value of: \(\rm \left[\frac{\cos 50^\circ}{1+\sin 50^\circ}\right]+\left[\frac{1+\sin 50^\circ}{\cos50^\circ}\right]\)

Formula Used:

cos² θ + sin² θ = 1

Calculation:

\(\rm \left[\frac{\cos 50^\circ}{1+\sin 50^\circ}\right]+\left[\frac{1+\sin 50^\circ}{\cos50^\circ}\right]\)

Take the LCM of the two terms:

\(\frac{cos² 50° + (1 + sin 50°)² }{(1 + sin 50°) × cos 50°}\)

Expanding the numerator:

cos² 50° + (1 + 2sin 50° + sin² 50°)

Using the identity cos² θ + sin² θ = 1:

1 + (1 + 2sin 50°)

2 + 2sin 50°

Now, the expression becomes:

\(\frac{2 + 2sin 50°}{(1 + sin 50°) × cos 50°}\)

Factor out 2 from the numerator:

\(\frac{2(1 + sin 50°)}{(1 + sin 50°) × cos 50°}\)

Cancel out the common term (1 + sin 50°):

2 / cos 50°

The value of 1 / cos 50° is sec 50°.

Therefore, the expression simplifies to: 2 sec 50°

∴ The value of the given expression is 2 sec 50°.

Given:

sin 54º

cos 72º

Formula Used:

sin (90º - θ) = cos θ

cos (90º - θ) = sin θ

Calculation:

Using the formula, we have:

sin 54º = cos (90º - 54º)

sin 54º = cos 36º

cos 72º = sin (90º - 72º)

cos 72º = sin 18º

So, sin 54º + cos 72º can be written as:

sin 54º + cos 72º = cos 36º + sin 18º

The correct answer is option 3: cos 36º + sin 18º

Given:

In a triangle PQR, ∠P + ∠Q = 84º.

Formula Used:

Sum of angles in a triangle = 180º

Calculation:

Let ∠R be the third angle of the triangle.

Sum of angles in a triangle = ∠P + ∠Q + ∠R

⇒ 180º = 84º + ∠R

⇒ ∠R = 180º - 84º

⇒ ∠R = 96º

The value of ∠R is 96º.

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

Calculation:

tan 4384° + cot 6814° 

⇒ tan (180° × 24 + 64°) + cot (90° × 75 + 64°)

⇒ tan  64°-  tan  64° = 0

∴ The correct option is 3

Concept:

sin(90 - a) = cos a

Cos(90 - a) = sin a

Calculation

sin 25° sin 65° – cos 25° cos 65°

⇒ sin 25° sin (90 - 25)° – cos 25° cos (90 - 25)°

⇒ sin 25° cos 25° – cos 25° sin 25°

⇒ 0

The value is 0.

Calculation:

As per the question,

 sin P + sin Q + sin R = QR/PR + 1 + PQ/PR

The sine of a 90-degree angle is always 1, because for a right angle, the opposite side and the hypotenuse are equal.

= 4/5 + 1 + 3/5

= 12/5

Given:

tan 40° = α 

Formula used:

Tan (A - B) = (tan A - tan B)/1 + tan A × tan B

tan (90° - θ) = cot θ 

cot θ × tan θ = 1

Calculation:

⇒ \(\frac{tan320^0 - tan 310^0}{1 + tan320^0.tan 310^0}\) = tan (320° - 310°)

⇒ tan 10° 

Now, we can write:

Tan 10° = tan (50° - 40°)

⇒ [(tan 50° - tan 40°)/1 + (tan 50° × tan 40°)]

⇒ [tan (90° - 40°) - tan 40°/1 + tan (90° - 40°) × tan 40°]

⇒ cot 40° - tan 40°/1 + cot 40° × tan 40°

⇒ (1/α - α)/1 + 1

⇒ (1/α - α)/2

⇒ (1 - α²)/2α 

∴ The correct answer is (1 - α2)/2α.

Concept used:

Sin²θ + Cos²θ = 1

Cos 2θ - Sin2θ = Cos2θ

Calculation:

\({\cos^215^{\circ} - \sin^215^{\circ}} \over{\cos^2145^{\circ} + \sin^2145^{\circ}}\)

⇒ Cos (2 ×15°)

⇒ Cos 30° = \(\sqrt3 \over 2\)

∴ The correct option is 3

Concept used:

sin3θ = 3sinθ  - 4 sin3θ

Calculation:

sin 10°- \(\dfrac{4}{3}\) sin3 10°

⇒ \(\dfrac{\text{3 Sin10 - 4Sin}^310}{3}\)

⇒ \(\dfrac{\text{Sin3(10)}}{3}\) = 

⇒ \(\dfrac{\text{Sin30}}{3}\) = \(\dfrac{1}{2}\) × \(\dfrac{1}{3}\) = \(\dfrac{1}{6}\)

sin 10°- \(\dfrac{4}{3}\)sin3 10°= \(\dfrac{1}{6}\)

Given:

AD ⊥ BC ; ∠D = 90° 

AC = 26 units; CD = 10 units; BC = 42 units

∠DAC = x and ∠B = y

Formula used:

Pythagorean theorem:

H² = P² + B²

Cos θ = B/H ; tan θ = P/B

Where, H = hypotenuse; P = perpendicular; B = Base

Calculation:

In △DAC

⇒ AC² = AD² + CD²

⇒ 26² = AD2 + 10²

⇒ 676 = AD2 + 100

⇒ AD = √(676 - 100) = √576 = 24 units

In △ADB

BD = (BC - CD) = (42 - 10) = 32 units

⇒ AB2 = AD2 + BD2

⇒ AB2 = 242 + 322

⇒ AB2 = 576 + 1024

⇒ AB = √1600 = 40 units

From the question:

\(\rm \frac{6}{\cos x}-\frac{5}{\cos y}+8\tan y\)

⇒ 6/(AD/AC) - 5/(BD/AB) + 8 × (AD/BD)

⇒ 6/(24/26) - 5/(32/40) + 8 × (24/32)

⇒ (6 × 26/24) - (5 × 40/32) + 8 × (24/32)

⇒ 6.5 - 6.25 + 6

⇒ 6.25 = 25/4 units

∴ The correct answer is 25/4 units.   

Calculation:

cosec 2910° + sec 4260° + tan 2565° + cot 1755°

⇒ cosec (2970 - 60)° + sec (4230 + 30)° + tan (2610 - 45)° + (cot 1800 - 45)° 

⇒ cosec (33 × 90 - 60)° + sec (47 × 90 + 30)° + tan (29 × 90 - 45)° + (cot 20 × 90 - 45)°

⇒ sec60° + cosec30° + cot45° - cot45°

⇒ 2 + 2 + 1 - 1 = 4

∴ The correct option is 3
 Alternate Method

cosec 2910° + sec 4260° + tan 2565° + cot 1755°

⇒ cosec (360 × 8 + 30)° + sec(360 × 12 - 60)° + tan(360 × 7 + 45)° + cot(360 × 5 - 45)°

⇒ cosec30° + sec60° + tan45° - cot45°

⇒ 2 + 2 + 1 - 1 = 4

∴ The correct option is 3.