WBBSE Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Multiple Choice Questions

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WBBSE Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Multiple Choice Questions

Example 1. If 3x = cosec α and \(\frac{3}{x}\) = cot α, then the value of 3\(\left(x^2-\frac{1}{x^2}\right)\) is

  1. \(\frac{1}{27}\)
  2. \(\frac{1}{81}\)
  3. \(\frac{1}{3}\)
  4. \(\frac{1}{9}\)

Solution: \((3 x)^2-\left(\frac{3}{x}\right)={cosec}^2 \alpha-\cot ^2 \alpha\)

Read And Learn Also WBBSE Class 10 Maths Multiple Choice Questions

⇒ \(9\left(x^2-\frac{1}{x^2}\right)=1 \quad \Rightarrow 3\left(x^2-\frac{1}{x^2}\right)=\frac{1}{3}\)

∴ The Correct Answer is 3. \(\frac{1}{3}\)

WBBSE Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Multiple Choice Questions

Example 2. If 2x = sec A and \(\frac{2}{x}\) = tan A, then the value of 2\(\left(x^2-\frac{1}{x^2}\right)\) is

  1. \(\frac{1}{2}\)
  2. \(\frac{1}{4}\)
  3. \(\frac{1}{8}\)
  4. \(\frac{1}{16}\)

Solution: \((2 x)^2-\left(\frac{2}{x}\right)^2=\sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}\)

⇒ \(4\left(x^2-\frac{1}{x^2}\right)=1\)

⇒ \(2\left(x^2-\frac{1}{x^2}\right)=\frac{1}{2}\)

∴ The correct answer is 1. \(\frac{1}{2}\)

Class 10 Maths Trigonometry Chapter 2 MCQs

Example 3. If tan α+ cot α = 2 then the value of (tan13 α + cot13 α) is

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

Solution: \(\tan \alpha+\cot \alpha=2\)

⇒ \(\tan \dot{\alpha}+\frac{1}{\tan \alpha}=2\)

⇒ \(\frac{\tan ^2 \alpha+1}{\tan \alpha}=2\)

⇒ \(\tan ^2 \alpha+1=2 \tan \alpha \Rightarrow \tan ^2 \alpha-2 \tan \alpha+1=0\)

⇒ \((\tan \alpha-1)^2=0\)

⇒ \(\tan \alpha-1=0 \quad \Rightarrow \quad \tan \alpha=1\)

∴ \(\cot \alpha=\frac{1}{\tan \alpha}=1\)

\(\tan ^{13} \alpha+\cot ^{13} \alpha\)

= \((1)^{13}+(1)^{13}=1+1=2\)

∴ The correct answer is 3. 2

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Example 4. If sin θ- cos θ = 0 [0° ≤ θ ≤ 90°] and sec θ + cosec θ = x, then the value of x is

  1. 1
  2. 2
  3. 2
  4. 2√2

Solution: sin θ – cos θ = 0

⇒ sin θ = cos θ

⇒ \(\frac{\sin \theta}{\cos \theta}=1\)

⇒ tan θ = 1 = tan 45° [0° ≤ θ ≤ 90°]

⇒ θ = 45°

x = sec θ + cosec θ = sec 45° + cosec 45°

= √2 + √2 = 2√2

∴ The correct answer is 4. 2√2

Heights And Distances Mcqs With Solutions Class 10

Example 5. If 2 cos 3θ = 1, then the value of θ is

  1. 10°
  2. 15°
  3. 20°
  4. 30°

Solution: 2 cos 3θ= 1

⇒ cos 3θ = \(\frac{1}{2}\)

⇒ cos 3θ= cos 60°

⇒ 3θ= 60° ⇒ θ = 20°

∴ The correct answer is 3. 20°

Example 6. If sin θ + cosec θ =2 then the value of (sin θ – cosec θ) is

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

Solution: sin θ + cosec θ = 2

(sin θ – cosec θ)2 = (sin θ+ cosec θ)2 – 4 sin θ.cosec θ

= (2)2 – 4 x 1 = 0

⇒ sin θ – cosec θ = 0

∴ The correct answer is 3. 0

Example 7. If tan θ = \(\frac{a}{b}\) then the value of sin θ is

  1. \(\frac{a}{\sqrt{a^2-b^2}}\)
  2. \(\frac{b}{\sqrt{a^2+b^2}}\)
  3. \(\frac{a+b}{\sqrt{a^2+b^2}}\)
  4. \(\frac{a}{\sqrt{a^2+b^2}}\)

Solution: tan θ = \(\frac{a}{b}\)

⇒  cot θ = \(\frac{b}{a}\)

cosec θ = \(\sqrt{1+\cot ^2 \theta}=\sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{a^2+b^2}}{a}\)

sin θ = \(\frac{a}{\sqrt{a^2+b^2}}\)

∴ The correct answer is 4. \(\frac{a}{\sqrt{a^2+b^2}}\)

Class 10 Trigonometry Chapter 2 Mcqs With Answers

Example 8. If sec θ – cosec θ = 0 then the value of tan θ is

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

Solution: sec θ – cosec θ = 0

⇒ sec θ = cosec θ

⇒ \(\frac{1}{\cos \theta}=\frac{1}{\sin \theta}\)

⇒ cos θ = sin θ

tan θ = \(\frac{\sin \theta}{\cos \theta}=\frac{\sin \theta}{\sin \theta}\) = 1

∴ The correct answer is 2. 1

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