Trigonometric Ratios Practice Questions (original) (raw)

Last Updated : 23 Jul, 2025

**Trigonometric Ratios are everyday concepts with real-life applications. Understanding them is essential for solving aptitude and reasoning questions.

This article offers a variety of easy-to-understand trigonometric ratios questions.

What are Trigonometric Ratios?

**Trigonometric Ratios are the basic relations between the sides of a right-angle triangle and their corresponding angles. These ratios are essential for solving problems involving triangles and modelling periodic phenomena.

The six trigonometric ratios are:

• Sine
• Cosine
• Tangent
• Cosecant
• Secant
• Cotangent

The mathematical symbol θ is used to denote the angle. Here θ =\angle ACB.

Triangle

**Note : To calculate these T-Ratios, you can follow a simple trick to find it. The trick is called as "**SOH-CAH-TOA" where "SOH" means Sine is Opposite/Hypotenuse, "CAH" means Cosine is Adjacent/Hypotenuse and "TOA" means Tangent is Opposite/adjacent. Remember that cosecant is 1/sine, secant is 1/cosine and cotangent is 1/tangent.

Important Formulas/Concepts

Follow the Trigonometric Ratios Tableadded below:

**Trigonometric Table of Some Specific Angles

Trigonometric Ratios for standard angles 0°, 30°, 45°, 60°, and 90º are required to solve trigonometric ratios.

Practice Questions on Trigonometric Ratios

**Question 1: Express 1.2 radians in degree measures.

**Solution:

To find: 1.2 radians in degrees measures [1 radian = (180/π) degrees]

1.2 radians = 1.2 × (180/π) degrees [Approximate the value of π as 22/7]

= 1.2 × (180/22/7)

= 1.2 × 180 × 7/22

= 216 × 7/22

= 68.72 degrees

**Therefore, 1.2 radians is approximately equal to 68.72 degrees.

**Question 2: If cosec A + cot A =11/2, then find the value of tan A.

**Solution:

Given that: cosec A + cot A =11/2 - (i)

Lets find what is 1/ (cosec A + cot A)

1/ (cosec A + cot A)

= 1/(1/sin A + cos A/sin A)

= 1/((1+cos A)/sin A)

= sin A / (1+cos A) [Multiplying and dividing by (1 - cos A)]

= sin A (1 - cos A) / [(1+cos A)(1-cos A)]

= sin A (1 - cos A) / (1-cos2A) [sin2A=(1-cos2A)]

= sin A (1 - cos A) / sin2A

= (1 - cos A) / sin A

= 1/sin A - cos A/sin A

= cosec A - cot A

From equation (i), we get

cosec A - cot A =2/11 - (ii)

Solving (i) - (ii), we get

2cot A = 117/22

tan A = 44/117

**Therefore, the value of tan A is equal to 44/117.

**Question 3: If cot A =1/2, then find the values of sin 2A and cos2A.

**Solution:

Given cot A = 1/2 - (i)

tan A = 1/ cot A = 2 - (ii)

Using the sin 2A, cos 2A formulas, (i) and (ii), we get

sin 2A = 2tanA/(1+tan2A)

= 2×2/(1+22) = 4/5

cos 2A = (1-tan2A)/(1+tan2A)

= (1-22)/(1+22) = -3/5

**Therefore,the value of sin 2A is equal to 4/5 and cos 2A is equal to -3/5.

**Question 4: Solve sin 6° sin 42° sin 66° sin 78°.

**Solution:

To find: sin 6° sin 42° sin 66° sin 78°

sin 6° sin 42° sin 66° sin 78° [sin(90 - A) = cos A]

= sin 6° cos 48° cos 24° cos 12° [Multiplying and dividing by 23 sin 12°]

= sin 6° (23 sin 12°) cos 12° cos 24° cos 48°/(23 sin 12°) [sin 2A = cos A sin A]

= 22sin 6° (2sin 12° cos 12°) cos 24° cos 48°/(23 sin 12°)

= 2sin 6° (2sin 24° cos 24°) cos 48°/(23 sin 12°)

= sin 6° (2sin 48° cos 48°)/(23 sin 12°)

= sin 6° sin 96°/(23 sin 12°)

= sin 6° cos 6°/(23 sin 12°)

= sin 12° /(2.23 sin 12°) = 1/24

= 1/16

**Therefore, the value of sin 6° sin 42° sin 66° sin 78° is equal to 1/16.

**Question 5: Solve sin 10° sin 30° sin 50° sin 70°.

**Solution:

To find: sin 10° sin 30° sin 50° sin 70° [sin A sin (60°-A) sin (60°+A) = sin 3A /4]

sin 10° sin 30° sin 50° sin 70°

= sin 10°sin (60°-10°) sin (60°+10°) sin 30°

= 1/4 sin230°

= 1/16

**Therefore, the value of sin 10° sin 30° sin 50° sin 70° is equal to 1/16.

**Question 6: If sin A = -1/√2 and tan A = 1 then A lies in which quadrant.

**Solution:

Given that, sin A = -1/√2 (negative)

tan A = 1 (positive)

According to CAST Rule, sin A is negative in 3rd and 4th quadrant whereas tan A is positive in 1st and 3rd quadrant.

**Therefore A lies in 3rd quadrant.

**Question 7: Find the value of sin (π +A)sin(π -A) cosec 2 A.

**Solution:

To find: sin (π +A)sin(π -A) cosec2A

sin (π +A)sin(π -A) cosec2A

= (- sin A)(sin A)cosec2A [cosec2A = 1/ sin2A]

=- sin2A/sin2A = -1

**Therefore, the value of sin (π +A)sin(π -A) cosec 2 A is equal to -1.

**Question 8: Find the value of tan1°tan2°tan3°........ tan 89°.

**Solution:

To find: tan1°tan2°tan3°........tan 89°

= tan1°tan2°tan3°........tan 89° [tan (90-A)=cot A]

= tan1°tan2°tan3°.......tan 45°........cot3°cot2°cot 1° [cot A tan A = 1 as cot A = 1/tan A]

= tan 45°

= 1

**Therefore, the value of tan1°tan2°tan3°........ tan 89° is equal to 1.

Practice Questions on Trigonometric Ratios: Unsolved

Practice Questions on Trigonometric Ratios : Unsolved

Practice Questions on Trigonometric Ratios: Unsolved

**You can download this practice Question from here: Practice Questions on Trigonometric Ratios: Unsolved

Answer Key:

• Ans 1. 0
• Ans 2. 1
• Ans 3. Thus, the solution to the given expression (tan 155°-tan 115°)/(1+tan 155°.tan 115°) is: \frac{1 - x^2}{2x}
• Ans 4. 1
• Ans 5. The angle A lies in **Quadrant III.
• Ans 6. SinA = \frac{\sqrt{3}}{2} \ \ \ and \ \ \ \ \ \frac{-\sqrt{3}}{2}
• Ans 7. The maximum and minimum value of cos 2x + cos2x are 2 and -1.
• Ans 8. The value of sin 27°-cos 27 = \frac{-\sqrt{2}\left( \sqrt{5} -1 {}\right)}{4}
• Ans 9. x=\frac{\left(-1 +\sqrt{5}{}\right)}{4} and \frac{\left(-1 -\sqrt{5}{}\right)}{4}
• Ans 10. \frac{1}{16}

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• Trigonometry in Maths
• Domain and Range of Trigonometric Functions