Trigonometric Ratios Practice Questions

Last Updated : 18 Jun, 2026

There are three sides of a triangle: Hypotenuse, Adjacent, and Opposite.

triang
  • The ratios between these sides based on the angle between them are called Trigonometric Ratios.
  • The six trigonometric ratios are sine (sin), cosine (cos), tangent (tan), cotangent (cot), cosecant (cosec), and secant (sec).

Question 1: Express 1.2 radians in degree measures.

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

1.2 radians = 1.2 Ɨ (180/Ļ€) degrees [Approximate the value of Ļ€ as 22/7]

1.2 Ɨ (180/22/7)

1.2 Ɨ 180 Ɨ 7/22

216 Ɨ 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.

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.

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)

= 2Ɨ2/(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°.

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°.

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.

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 (Ļ€ +A)sin(Ļ€ -A) cosec2A.

To find: sin (Ļ€ +A)sin(Ļ€ -A) cosec2A

sin (Ļ€ +A)sin(Ļ€ -A) cosec2A

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

- sin2A/sin2A = -1

Therefore, the value of sin (Ļ€ +A)sin(Ļ€ -A) cosec2A is equal to -1.

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

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 : Unsolved

Q1. Find the value of cos⁔1∘cos⁔2∘cos⁔3∘ ........cos⁔179∘.

Q2. If a+b=Ļ€/2 and b+c=a, then find tan⁔ a.

Q3. If tan⁔25∘=x, then find \frac{\tan 155^\circ-\tan 115^\circ}{1+\tan 155^\circ \tan 115^\circ}

Q4. If A+B+C=3Ļ€/ 2, then find cos⁔2A + cos⁔2B + cos⁔2C.

Q5. Predict the quadrant where A lies, if sin⁔A is negative and tan⁔A is positive.

Q6. Calculate sin⁔A, if cos⁔A=1/2

Q7. Find the maximum and minimum values of cos⁔2x+cos⁔2x.

Q8. Find the value of sin⁔27āˆ˜āˆ’cos⁔27∘.

Q9. If tan⁔1∘tan⁔2∘tan⁔3āˆ˜ā‹Ætan⁔89∘=4x2+2x = 4x2+2x, then find the value of x.

Q10. Find the value of \cos\frac{\pi}{10}\, \cos\frac{2\pi}{10}\, \cos\frac{4\pi}{10}\, \cos\frac{8\pi}{10}\, \cos\frac{16\pi}{10}

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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