WBBSE Class 10 Maths Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Multiple Choice Questions

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WBBSE Class 10 Maths Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Multiple Choice Questions

Example 1. The value of (sin 43° cos 47° + cos 43° sin 47°) is

  1. 0
  2. 1
  3. sin 4°
  4. cos 4°

Solution: sin 43° cos 47° + cos 43° sin 47°

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

⇔ sin 43° cos (90° – 43°) + cos 43° sin (90° – 43°)

= sin 43° sin 43° + cos 43° cos 43°

= sin2 43° + cos2 43° – 1

∴ The correct answer is 2. 1

WBBSE Class 10 Maths Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Multiple Choice Questions

Example 2. The value of \(\left(\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 78^{\circ}}{\tan 12^{\circ}}\right)\) is

  1. 0
  2. 1
  3. 2
  4. None of these

Solution: \(\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 78^{\circ}}{\tan 12^{\circ}}\)

= \(\frac{\tan 35^{\circ}}{\cot \left(90^{\circ}-35^{\circ}\right)}+\frac{\cot 78^{\circ}}{\tan \left(90^{\circ}-78^{\circ}\right)}=\frac{\tan 35^{\circ}}{\tan 35^{\circ}}+\frac{\cot 78^{\circ}}{\cot 78^{\circ}}\)

= 1 + 1 = 2

∴ The correct answer is 3. 2

Class 10 Trigonometry Chapter 3 Mcqs With Answers

Example 3. The value of {cos (40° + θ) – sin (50° – θ)} is

  1. 2 cosθ
  2. 7 sinθ
  3. 0
  4. 1

Solution: cos (40° + θ) – sin (50° – θ)

= cos (40° +θ) – cos {90° – (50° – θ)) = cos (40° + θ) – cos (90° – 50° + θ)

= cos (40° + θ) – cos (40° + θ) = 0

∴ The correct answer is 3. 0

Example 4. ABC is a triangle, \(\sin \left(\frac{B+C}{2}\right)\) =

  1. sin \(\frac{A}{2}\)
  2. cos \(\frac{A}{2}\)
  3. sin A
  4. cos A

Solution: A + B + C = 180°

\(\sin \left(\frac{B+C}{2}\right)=\sin \left(\frac{180^{\circ}-A}{2}\right)\)

= \(\sin \left(90^{\circ}-\frac{A}{2}\right)=\cos \frac{A}{2}\)

∴ The correct answer is 2. cos \(\frac{A}{2}\)

Example 5. If A + B = 90° and tan A = \(\frac{3}{4}\), then the value of cot B is

  1. \(\frac{3}{4}\)
  2. \(\frac{4}{3}\)
  3. \(\frac{3}{5}\)
  4. \(\frac{4}{5}\)

Solution: tan A = \(\frac{3}{4}\)

⇒ tan (90° – B) = \(\frac{3}{4}\) [as A + B = 90°]

⇒ cot B = \(\frac{3}{4}\)

∴ The correct answer is 1. \(\frac{3}{4}\)

Class 10 Maths Chapter 3 Trigonometry Solutions 

Example 6. If tanθ tan 2θ = 1, and 20 is positive acute angle then the value of sin 3θ is

  1. 0
  2. 1
  3. -1
  4. 2

Solution: tanθ tan 2θ= 1

⇒ tan 2θ = \(\frac{1}{\tan \theta}\)

⇒ tan 2θ = cotθ

⇒ tan 2θ = tan (90° – θ)

⇒ 2θ = 90° – θ

⇒ 3θ = 90°

sin 3θ = sin 90° = 1

∴ The correct answer is 2. 1

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Example 7. The value of \(\left(\frac{1}{\cos ^2 75^{\circ}}-\frac{1}{\tan ^2 15^{\circ}}\right)\) is

  1. 1
  2. 0
  3. -1
  4. 2

Solution: \(\left(\frac{1}{\cos ^2 75^{\circ}}-\frac{1}{\tan ^2 15^{\circ}}\right)\)

= sec275° – cot215°

= sec275° – cot2 (90° – 75°)

= sec275° – tan275° = 1

∴ The correct answer is 1. 1

Trigonometric Identities Mcqs With Solutions Class 10

Example 8. If α + β = 90°, then the value of \(\frac{\tan \alpha-\cot \beta}{\tan \alpha+\cot \beta}\) is

  1. 3
  2. 1
  3. 0
  4. \(\frac{1}{\sqrt{3}}\)

Solution: \(\frac{\tan \alpha-\cot \beta}{\tan \alpha+\cot \beta}=\frac{\tan \alpha-\cot \left(90^{\circ}-\alpha\right)}{\tan \alpha+\cot \left(90^{\circ}-\alpha\right)}\)

= \(\frac{\tan \alpha-\tan \alpha}{\tan \alpha+\tan \alpha}=0\)

∴ The correct answer is 3. 0

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