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

Calculation: 

We are given:

We need to find in terms of p  and q .

We start by simplifying the expression

This simplifies further as:

Now, simplify the fraction inside the denominator:

Now, expand both terms using trigonometric identities:

Simplifying this expression gives:

∴ The correct answer is Option (c):

Calculation: 

We are given:

We need to find p + q.

Simplifying both terms:

Now, factorizing and simplifying further:

Recognizing the sine identity, we get:

Finally, simplifying this:

The final result is:

∴ The correct answer is Option (4)

Explanation:

We are given:

We rewrite q  in terms of tangent since 

Now we compute

Next, we find a common denominator for q :

Substituting this back into the formula for

Simplifying, we get:

∴ The correct answer is Option (c):

Concept:

sin2x + cos2x = 1

Calculation:

Given: sin x + sin² x = 1

As we know that, sin2x + cos2x = 1

⇒ sin x + (1 – cos² x) = 1

⇒ sin x = cos² x ----(1)

⇒ sin² x = cos⁴ x (Using (1)) 

⇒ cos² x + cos⁴ x = sin x + sin² x = 1

Given:

(1 + cos²A) = 3sinA.cosA

Solution:

Dividing by cos²A,

⇒ (1 + cos2A) = 3sinA.cosA

⇒ (1 + sec²A) = 3tanA

We know that: sec²A = (1 + tan²A)

⇒ 2 + tan²A = 3tanA

⇒ tan²A – 3tanA + 2 = 0

⇒ (tan A – 1) (tan A – 2) = 0

⇒ tan A = 1 or 2

Calculation:

Given tan θ = .

Consider a right angled triangle with perpendicular 4 units and base 3 units.

By pythagorus theorem h = 

⇒ h = 5.

∴ sin θ = 

⇒ sin θ = 

We know that tan function is negative in 2nd and 4th quadrant.

Sin function is positive in 2nd quadrant and negative in 4th quadrant.

If θ is in 2nd quadrant, sin θ  = .

If θ is in 4th quadrant, sin θ = .

sin θ can be  or .

Concept:

• sin² x + cos² x = 1
• cos (A + B) = cos A cos B – sin A sin B
• cos (A – B) = cos A cos B + sin A sin B

Calculation:

Given:  and θ is in 1^st quadrant

cos (30 + θ) = cos 30 cos θ – sin 30 sin θ

     ----(1)

cos (45 – θ) = cos 45 cos θ + sin 45 sin θ

      ----(2)

cos (120 – θ) = cos 120 cos θ + sin 120 sin θ

 ----(3)

Adding (1), (2) and (3)

Given

(cosec x + cot x)/(cosec x – cot x) = 7

Formula used

Cos²x = 1 – sin²x

Calculation

(cosec x + cot x)/(cosec x – cot x) = 7

⇒ (cosec x + cot x = 7(cosec x – cot x)

⇒ 8cot x = 6 cosec x

⇒ (8cos x)/sin x = 6/sin x

⇒ Cos x = 3/4

⇒ Cos² x = 9/16

⇒ Sin² x = 1 – 9/16

⇒ Sin² x = 7/16

⇒ (4sin2x + 1)/(4sin2x – 1) = (4 × 7/16 + 1)/(4 × 7/16 – 1)

⇒ (7/4 + 1)/(7/4 –1)

⇒ (11/4)/(3/4)

⇒ 11/3

∴ The value of  (4sin2x + 1)/(4sin2x – 1) is 11/3

∴ sin² θ + cos² θ = 1

Concept:

sec^-1 x = cos^-1 (1/x)

cosec^-1 (x) + sec^-1 (x) = π / 2

Calculation:

As we know that, 

sec-1 x = cos-1 (1/x)

As we know that,

cosec-1 (x) + sec-1 (x) = π / 2

Given:

tan 

⇒ tan A = 11/60

Formulas used:

tan A = Perpendicular/Base, where A is an acute angle

Pythagoras Theorem 

In triangle xyz, xz = Hypotenuse

(xz)² = (xy)² + (yz)²

Calculation:

(xz)² = 11² + 60² 

⇒ xz = √3721 = 61 

Now,  (4cos A - 7sin A) 

⇒ 4 × B/H - 7 × P/H 

⇒ 4 × 60/61 - 7 × 11/61 

⇒ 240/61 - 77/61 = 163/61 

⇒ 

∴ (4cos A - 7sin A) =  

Concept:

• sin 60° = 

• cos 30° = 

Calculation:

Since, sin (A + B) = 

⇒ sin (A + B) = sin 60° 

⇒ A + B = 60°      ----(1)

Also, cos (A - B) = 

⇒ cos (A - B) = cos 30° 

⇒ A - B = 30°      ----(2)

On solving equations (1) & (2), we get,

A = 45° and B = 15° 

Hence, acute angles of A and B are 45° and 15° respectively.

Given:

We have to evalute 2 tan2 45° + cos2 30° - sin2 60°

Formula Used:

tan45° = 1 

Cos30° = √3/2

Sin60° = √3/2

Calculation:

2 tan² 45° + cos² 30° - sin² 60°

⇒ 2 × 1² + (√3/2)² – (√3/2)²

⇒ 2 + 3/4 – 3/4

⇒ 2

∴ The value of 2 tan2 45° + cos2 30° - sin2 60° is 2.