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

Given:

If sec 4A = cosec (3A - 50°), where 4A and 3A are acute angles, find the value of cosec (A + 25º).

Formula Used:

We know that sec θ = cosec (90° - θ).

Calculation:

Given sec 4A = cosec (3A - 50°)

 ⇒ 4A = 90° - (3A - 50°)

⇒ 4A = 90° - 3A + 50°

⇒ 4A + 3A = 140°

⇒ 7A = 140°

⇒ A = 20°

Now, we need to find A + 25:

⇒ A + 25 = 20° + 25°

⇒ A + 25 = 45°

cosec (A + 25º) = cosec 45° = √2

∴ The correct answer is √2.

We are given the equation:

1 / (cosec² θ - cot² θ) + 1 / (sec² θ - tan² θ) = 2 sin θ

Let's simplify each term:

We know that:

cosec² θ - cot² θ = 1 (using the identity cosec² θ - cot² θ = 1)

sec² θ - tan² θ = 1 (using the identity sec² θ - tan² θ = 1)

So, we can rewrite the expression as:

1 / 1 + 1 / 1 = 2

This simplifies to:

2 = 2 sin θ

Now, divide both sides by 2:

1 = sin θ

Therefore, one of the values of θ that satisfies this equation is:

θ = 90°

The value of θ is:  90°

Given:

sin A = 3/4

Formula used:

sin²A + cos²A = 1

Calculations:

(3/4)² + cos²A = 1

⇒ 9/16 + cos²A = 1

⇒ cos²A = 1 - 9/16

⇒ cos²A = 16/16 - 9/16

⇒ cos²A = 7/16

⇒ cos A = √7/4

∴ The value of cos A is √7/4.

Given:

A mirror is placed on the ground facing upwards. A man sees the top of a tower in the mirror which is at a distance of 105 m from the mirror. The man is 0.5 m away from the mirror, and his height is 1.5 m.

Calculation:

Let AB and DE be the man and Tower respectively.

In Δ ABC and Δ CDE are similar 

So,

DE / CD = AB / BC

⇒ h / 105 = 1.5 / 0.5

⇒ h = 105 × 3 = 315 m

∴ The correct answer is option 2.

Given:

Height of the wall = 12 metres

Length of the ladder = 13 metres

Calculations:

Let the distance of the base of the ladder from the wall be d metres.

Using the Pythagorean theorem (a² + b² = c²):

(distance from the wall)² + (height of the wall)² = (length of the ladder)²

d² + 12² = 13²

d² + 144 = 169

d² = 169 - 144

d² = 25

d = √25

d = 5

∴ The base of the ladder is at a distance of 5 metres from the wall on the ground.

GIVEN:

BC = 18 m

CONCEPT:

FORMULAE USED:

Tanθ = Perpendicular/Base

Cosθ = Base/Hypotenuse

CALCULATION:

Height of the tree = AB + AC

Tan 30° = AB/18

⇒ (1/√3) = AB/18

⇒ AB = (18/√3)

Cos 30° = BC/AC = 18/AC

⇒ √3/2 = 18/AC

⇒ AC = 36/√3

Hence, AB + AC = 18/√3 + 36/√3 = 54 / √3

⇒ 54/√3 × √3 /√3  (rationalizing to remove root from denominator)

⇒ 54√3 / 3 = 18√3

∴ Height of the tree = 18√3.

Mistake Point: Here, total height of tree is (AB + AC).

The above Question is previous year Question taken directly from NCERT class 10th. Correct answer will be 18√3

We can find the angle of elevation by using the following steps:

Calculation:

Label the height difference between the two points on the ground as "h" and the horizontal distance between the two points as "d".

Use the tangent function to find the angle of elevation:

tan(θ) = \(\frac{h}{d}\).

Solve for the angle of elevation:

\(θ = tan^-1(\frac{h}{d}).\)

In this case, h = 20 m and d = 20√3 m, so:

\(tan(θ) = \frac{20 }{ (20√3)}\)

\(tan(θ) = \frac{1 }{ √3}\)

\(θ = tan^-1(\frac{1}{ √3})\)
θ = 30°

So the angle of elevation is 30°.

Given:

tan 53° = 4/3

Formula Used:

tan(x – y) = (tanx – tany)/(1 + tanxtany)

Calculation:

We know, 8° = 53° - 45°

Tan8° = tan(53° - 45°)

⇒ tan8° = (tan53° - tan45°)/(1 + tan53° tan45°)

⇒ tan8° = (4/3 – 1)/(1 + 4/3 × 1)

⇒ tan8° = (1/3)/(7/3)

Concept used: 

sec²(x) = 1 + tan²(x)

Calculation:

⇒ sec²θ + tan²θ = 5/3

⇒ 1 + tan²θ + tan²θ = 5/3

⇒ 2tan²θ = 2/3

⇒ tanθ = 1/√3

⇒ θ = 30

Given:

tanθ + cotθ = √3

Formula used:

(a + b)³ = a³ + b³ + 3ab(a + b)

a² + b² = (a + b)² - 2(a × b)

tanθ × cotθ = 1

Calculation:

tanθ + cotθ = √3

Taking cube on both sides, we get

(tanθ + cotθ)³ = (√3)³

⇒ tan³θ + cot³θ + 3 × tanθ × cotθ × (tanθ + cotθ) = 3√3

⇒ tan3θ + cot3θ + 3√3  = 3√3

⇒ tan3θ + cot3θ = 0  

Taking square on the both sides

(tan3θ + cot3θ)² = 0

⇒ tan⁶θ + cot⁶θ + 2 × tan3θ × cot3θ = 0

⇒ tan6θ + cot6θ + 2 = 0    

⇒ tan6θ + cot6θ = - 2

∴ The value of tan6θ + cot6θ is - 2.

As,

⇒ sec²θ = 1 + tan²θ

We have,

⇒ (sec²θ)² – sec²θ = 3

⇒ (1 + tan²θ)² – (1 + tan²θ) = 3

⇒ (1 + tan⁴θ + 2tan²θ) – (1 + tan²θ) = 3

⇒ 1 + tan⁴θ + 2tan²θ – 1 – tan²θ = 3

Formula Used:

1/Cosec Ø = Sin Ø 

Sin²Ø + Cos²Ø = 1

Calculation:

Cos2Ø + 1/Cosec2Ø + 17 = x

⇒ Cos2Ø + Sin2Ø + 17 = x

⇒ 1 + 17 = x

⇒ x = 18

⇒ x² = 324

∴ The value of x² is 324.

Given:

The given expression = \(\frac{{\sin 23^\circ \cos 67^\circ + \sec52^\circ \sin38^\circ + \cos 23^\circ \sin 67^\circ + \rm cosec52^\circ \cos 38^\circ }}{{\rm cose{c^2}20^\circ - {{\tan }^2}70^\circ }}\)

Formula used:

cos 67° = sin (90° - 67°) = sin 23°

sin 67° = cos (90° - 67°) = cos 23°

sin² θ + cos² θ = 1

sec² θ - tan² θ = 1

Calculation:

\(\frac{{\sin 23^\circ \cos 67^\circ + \sec52^\circ \sin38^\circ + \cos 23^\circ \sin 67^\circ + \rm cosec52^\circ \cos 38^\circ }}{{\rm cose{c^2}20^\circ - {{\tan }^2}70^\circ }}\)

= \(\frac{{\sin 23^\circ \sin 23^\circ +\frac{1}{cos\,52^\circ} \sin38^\circ + cos 23^\circ cos 23^\circ + \frac{1}{sin\,52^\circ} \cos 38^\circ }}{sec^2\,70^\circ-tan^2\,70^\circ}\)

= \(\frac{sin^2\,23^\circ+\frac{sin\,38^\circ}{sin\,38^\circ}+cos^2\,23^\circ+\frac{cos\,38^\circ}{cos\,38^\circ}}{1}\)

= sin2 23° + 1 + cos2 23° + 1

= 1 + 1 + 1

= 3

∴ The value of the given expression is 3

Calculation:

cot⁴θ + cot²θ = 3

⇒ cos⁴θ/sin⁴θ + cos ²θ/sin ²θ 

⇒ cos 2θ(cos 2θ+ sin 2θ )/sin4θ = 3 (Taking LCM)

⇒ cos 2θ/sin⁴θ = 3

⇒ cot²θ . cosec²θ = 3

Now, cosec⁴θ – cosec²θ

⇒ cosec²θ(cosec²θ – 1)

⇒ cosec²θcot²θ = 3

∴ cosec4θ – cosec2θ = 3

Given:

secθ - cosθ = 14 and 14 secθ = x

Concept used:

\(Sec\theta =\frac{1}{Cos\theta}\)

Calculations:

According to the question,

⇒ \(sec\theta - cos\theta= 14\)

⇒ \(\sec\theta-\frac{1}{sec\theta}=14\)

⇒ \( sec²\theta-1=14sec\theta\)

⇒ \(\tan^2\theta=14sec\theta\)      ----(\(sec²\theta-1=tan^2\theta\))

\(\ tan²\theta=x\)

∴ The value of x is \(tan²\theta\).