Class 10 RD Sharma Solutions – Chapter 6 Trigonometric Identities – Exercise 6.2

Last Updated : 28 Apr, 2021

Question 1. If cos θ = 4/5, find all other trigonometric ratios of angle θ.

Solution:

We are given, cos θ = 4/5. So, sec θ =1/cos θ = 5/4.

Now we know,

=> sin θ = √(1 – cos² θ)

=> sin θ = √(1 – (4/5)²)

=> sin θ = √(1 – (16/25))

=> sin θ = √(9/25)

=> sin θ = 3/5

So, cosec θ = 1/sin θ = 5/3

And tan θ = sin θ/cos θ = (3/5)/(4/5) = 3/4

Therefore, cot θ = 1/tan θ = 4/3

If cos θ = 4/5, value of sec θ, sin θ, cosec θ, tan θ and cot θ are 5/4, 3/5, 5/3, 3/4 and 4/3 respectively.

Question 2. If sin θ = 1/√2, find all other trigonometric ratios of angle θ.

Solution:

We are given, sin θ = 1/√2. So, cosec θ =1/sin θ = √2.

Now we know,

=> cos θ = √(1 – sin² θ)

=> cos θ = √(1 – (1/√2)²)

=> cos θ = √(1 – (1/2))

=> cos θ = √(1/2)

=> cos θ = 1/√2

So, sec θ = 1/cos θ = √2

And tan θ = sin θ/cos θ = (1/√2)/(1/√2) = 1

Therefore, cot θ = 1/tan θ = 1

If sin θ = 1/√2, value of cosec θ, cos θ, sec θ, tan θ and cot θ are √2, 1/√2, √2, 1 and 1 respectively.

Question 3. If tan θ = 1/√2, find the value of $\frac{cosec^2θ-sec^2θ}{cosec^2θ+cot^2θ}$ .

Solution:

We are given, tan θ = 1/√2. Now we know,

=> sec θ = √(1 + tan² θ)

=> sec θ = √(1 + (1/√2)²)

=> sec θ = √(1+(1/2))

=> sec θ = √(3/2)

And cot θ = 1/tan θ = √2. Also, we know,

=> cosec θ = √(1 + cot² θ)

=> cosec θ = √(1 + (√2)²)

=> cosec θ = √(1 + 2)

=> cosec θ = √3

So, $\frac{cosec^2θ-sec^2θ}{cosec^2θ+cot^2θ}$ = $\frac{(\sqrt{3})^2-\left(\sqrt{\frac{3}{2}}\right)^2}{(\sqrt{3})^2+\left(\sqrt{2}\right)^2}$

= $\frac{3-\frac{3}{2}}{3+2}$

= $\frac{3}{10}$

Therefore, the value of $\frac{cosec^2θ-sec^2θ}{cosec^2θ+cot^2θ}$ is $\frac{3}{10}$ .

Question 4. If tan θ = 3/4, find the value of $\frac{1-cosθ}{1+cosθ}$ .

Solution:

We are given, tan θ = 3/4. Now we know,

=> sec θ = √(1 + tan² θ)

=> sec θ = √(1 + (3/4)²)

=> sec θ = √(1+(9/16))

=> sec θ = √(25/16)

=> sec θ = 5/4

And cos θ = 1/sec θ = 4/5.

So, $\frac{1-cosθ}{1+cosθ}$ = $\frac{1-\frac{4}{5}}{1+\frac{4}{5}}$

= $\frac{\frac{1}{5}}{\frac{9}{5}}$

= $\frac{1}{9}$

Therefore, the value of $\frac{1-cosθ}{1+cosθ}$ is $\frac{1}{9}$ .

Question 5. If tan θ = 12/5, find the value of $\frac{1+sinθ}{1-sinθ}$ .

Solution:

We are given, tan θ = 12/5. So, cot θ = 1/tan θ = 5/12.

Now we know,

=> cosec θ = √(1 + cot² θ)

=> cosec θ = √(1 + (5/12)²)

=> cosec θ = √(1 + (25/144))

=> cosec θ = √(169/144)

=> cosec θ = 13/12

And sin θ = 1/cosec θ = 12/13.

So, $\frac{1+sinθ}{1-sinθ}$ = $\frac{1+\frac{12}{13}}{1-\frac{12}{13}}$

= $\frac{\frac{25}{13}}{\frac{1}{13}}$

= 25

Therefore, the value of $\frac{1+sinθ}{1-sinθ}$ is 25.

Question 6. If cot θ = 1/√3, find the value of $\frac{1-cos^2θ}{2-sin^2θ}$ .

Solution:

We are given, cot θ = 1/√3. Now we know,

=> cosec θ = √(1 + cot² θ)

=> cosec θ = √(1 + (1/√3)²)

=> cosec θ = √(1 + (1/3))

=> cosec θ = √(4/3)

=> cosec θ = 2/√3

And sin θ = 1/cosec θ = √3/2. Also, we know,

=> cos θ = √(1 – sin² θ)

=> cos θ = √(1 – (√3/2)²)

=> cos θ = √(1 – (3/4))

=> cos θ = √(1/4)

=> cos θ = 1/2

So, $\frac{1-cos^2θ}{2-sin^2θ}$ = $\frac{1-(\frac{1}{2})^2}{2-(\frac{\sqrt{3}}{2})^2}$

= $\frac{1-\frac{1}{4}}{2-\frac{3}{4}}$

= $\frac{\frac{3}{4}}{\frac{5}{4}}$

= $\frac{3}{5}$

Therefore, the value of $\frac{1-cos^2θ}{2-sin^2θ}$ is $\frac{3}{5}$ .

Question 7. If cosec A = √2, find the value of $\frac{2sin^2A+3cot^2A}{4(tan^2A-cos^2A)}$ .

Solution:

We are given, cosec A = √2. So, sin A = 1/cosec A = 1/√2.

Now we know,

=> cos A = √(1 – sin² A)

=> cos A = √(1 – (1/√2)²)

=> cos A = √(1 – (1/2))

=> cos A = √(1/2)

=> cos A = 1/√2

Hence, tan A = sin A/cos A = (1/√2)/(1/√2) = 1. And cot A = 1/tan A = 1.

So, $\frac{2sin^2A+3cot^2A}{4(tan^2A-cos^2A)}$ = $\frac{2(\frac{1}{\sqrt{2}})^2+3(1)^2}{4\left(1^2-(\frac{1}{\sqrt{2}})^2\right)}$

= $\frac{1+3}{2}$

= 2

Therefore, the value of $\frac{2sin^2A+3cot^2A}{4(tan^2A-cos^2A)}$ is 2.

Question 8. If cot θ = √3, find the value of $\frac{cosec^2θ+cot^2θ}{cosec^2θ-sec^2θ}$ .

Solution:

We are given cot θ = √3. And tan θ = 1/cot θ = 1/√3.

Now we know,

=> cosec θ = √(1 + cot² θ)

=> cosec θ = √(1 + (√3)²)

=> cosec θ = √4

=> cosec θ = 2

Also, we know,

=> sec θ = √(1 + tan² θ)

=> sec θ = √(1 + (1/√3)²)

=> sec θ = √(1+(1/3))

=> sec θ = √(4/3)

=> sec θ = 2/√3

So, $\frac{cosec^2θ+cot^2θ}{cosec^2θ-sec^2θ}$ = $\frac{2^2+(\sqrt{3})^2}{2^2-\left(\frac{2}{\sqrt{3}}\right)^2}$

= $\frac{7}{\frac{8}{3}}$

= $\frac{21}{8}$

Therefore, the value of $\frac{cosec^2θ+cot^2θ}{cosec^2θ-sec^2θ}$ is $\frac{21}{8}$ .

Question 9. If 3cos θ = 1, find the value of $\frac{6sin^2θ+tan^2θ}{4cosθ}$ .

Solution:

We are given cos θ = 1/3. Now we know,

=> sin θ = √(1 – cos² θ)

=> sin θ = √(1 – (1/3)²)

=> sin θ = √(1 – (1/9))

=> sin θ = √(8/9)

=> sin θ = 2√2/3

Hence, tan θ = sin θ/cos θ = (2√2/3)/(1/3) = 2√2

So, $\frac{6sin^2θ+tan^2θ}{4cosθ}$ = $\frac{6(\frac{2\sqrt{2}}{3})^2+(2\sqrt{2})^2}{4(\frac{1}{3})}$

= $\frac{\frac{16}{3}+8}{\frac{4}{3}}$

= $\frac{40}{4}$

= 10

Therefore, the value of $\frac{6sin^2θ+tan^2θ}{4cosθ}$ is 10.

Question 10. If √3 tan θ = sin θ, find the value of sin² θ – cos² θ.

Solution:

We are given, √3 tan θ = sin θ

=> √3 (sin θ/cos θ) = sin θ

=> cos θ = 1/√3

Now we know,

=> sin θ = √(1 – cos² θ)

=> sin θ = √(1 – (1/√3)²)

=> sin θ = √(1 – (1/3))

=> sin θ = √(2/3)

So, sin² θ – cos² θ = √(2/3)² – (1/√3)²

= 2/3 – 1/3

= 1/3

Therefore, the value of sin² θ – cos² θ is 1/3.

Question 11. If cosec θ = 13/12, find the value of $\frac{2sinθ-3cosθ}{4sinθ-9cosθ}$ .

Solution:

We are given, cosec θ = 13/12. So, sin θ = 1/cosec θ = 12/13.

Now we know,

=> cos θ = √(1 – sin² θ)

=> cos θ = √(1 – (12/13)²)

=> cos θ = √(1 – (144/169))

=> cos θ = √(25/169)

=> cos θ = 5/13

So, $\frac{2sinθ-3cosθ}{4sinθ-9cosθ}$ = $\frac{2(\frac{12}{13})-3(\frac{5}{13})}{4(\frac{12}{13})-9(\frac{5}{13})}$

= $\frac{\frac{24-15}{13}}{\frac{48-45}{13}}$

= $\frac{9}{3}$

= 3

Therefore, the value of $\frac{2sinθ-3cosθ}{4sinθ-9cosθ}$ is 3.

Question 12. If sin θ + cos θ = √2 cos (90^o–θ), find cot θ.

Solution:

We are given,

=> sin θ + cos θ = √2 cos (90^o–θ)

=> sin θ + cos θ = √2 sin θ

=> cos θ = (√2–1)sin θ

=> cos θ/sin θ = √2–1

=> cot θ = √2–1

Therefore, value of cot θ is √2–1.

Question 13. If 2sin² θ – cos² θ = 2, then find the value of θ.

Solution:

We have,

=> 2sin² θ – cos²θ = 2

=> 2sin² θ – (1 – sin² θ) = 2

=> 2sin² θ – 1 + sin² θ = 2

=> 3sin² θ = 3

=> sin² θ = 1

=> sin θ = 1

=> θ = 90^o

Therefore, the value of θ is 90^o.

Question 14. If √3tan θ – 1 = 0, find the value of sin² θ – cos² θ.

Solution:

We are given,

=> √3tan θ – 1 = 0

=> tan θ = 1/√3

=> tan θ = tan 30^o

=> θ = 30^o

So, sin² θ – cos² θ = sin² 30^o – cos² 30^o

= (1/2)² – (√3/2)²

= (1/4) – (3/4)

= –2/4

= –1/2

Therefore, the value of sin² θ – cos² θ is –1/2.

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Chapter 1: Real Numbers

Chapter 2: Polynomials

Chapter 3: Pair of Linear Equations in Two Variables

Chapter 4: Triangles

Chapter 5: Trigonometric Ratios

Chapter 6: Trigonometric Identities

Chapter 7: Statistics

Chapter 8: Quadratic Equations

Chapter 9: Arithmetic Progressions

Chapter 10: Circles

Chapter 11: Constructions

Chapter 12: Some Applications of Trigonometry

Chapter 13: Probability

Chapter 14: Coordinate Geometry

Chapter 15: Areas Related To Circles

Chapter 16: Surface Areas And Volumes

Chapter 1: Real Numbers

Chapter 2: Polynomials

Chapter 3: Pair of Linear Equations in Two Variables

Chapter 4: Triangles

Chapter 5: Trigonometric Ratios

Chapter 6: Trigonometric Identities

Chapter 7: Statistics

Chapter 8: Quadratic Equations

Chapter 9: Arithmetic Progressions

Chapter 10: Circles

Chapter 11: Constructions

Chapter 12: Some Applications of Trigonometry

Chapter 13: Probability

Chapter 14: Coordinate Geometry

Chapter 15: Areas Related To Circles

Chapter 16: Surface Areas And Volumes

Class 10 RD Sharma Solutions – Chapter 6 Trigonometric Identities – Exercise 6.2

Question 1. If cos θ = 4/5, find all other trigonometric ratios of angle θ.

Question 2. If sin θ = 1/√2, find all other trigonometric ratios of angle θ.

Question 3. If tan θ = 1/√2, find the value of .

Question 4. If tan θ = 3/4, find the value of .

Question 5. If tan θ = 12/5, find the value of .

Question 6. If cot θ = 1/√3, find the value of .

Question 7. If cosec A = √2, find the value of .

Question 8. If cot θ = √3, find the value of .

Question 9. If 3cos θ = 1, find the value of .

Question 10. If √3 tan θ = sin θ, find the value of sin2 θ – cos2 θ.

Question 11. If cosec θ = 13/12, find the value of .

Question 12. If sin θ + cos θ = √2 cos (90o–θ), find cot θ.

Question 13. If 2sin2 θ – cos2 θ = 2, then find the value of θ.

Question 14. If √3tan θ – 1 = 0, find the value of sin2 θ – cos2 θ.

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Question 3. If tan θ = 1/√2, find the value of $\frac{cosec^2θ-sec^2θ}{cosec^2θ+cot^2θ}$ .

Question 4. If tan θ = 3/4, find the value of $\frac{1-cosθ}{1+cosθ}$ .

Question 5. If tan θ = 12/5, find the value of $\frac{1+sinθ}{1-sinθ}$ .

Question 6. If cot θ = 1/√3, find the value of $\frac{1-cos^2θ}{2-sin^2θ}$ .

Question 7. If cosec A = √2, find the value of $\frac{2sin^2A+3cot^2A}{4(tan^2A-cos^2A)}$ .

Question 8. If cot θ = √3, find the value of $\frac{cosec^2θ+cot^2θ}{cosec^2θ-sec^2θ}$ .

Question 9. If 3cos θ = 1, find the value of $\frac{6sin^2θ+tan^2θ}{4cosθ}$ .

Question 10. If √3 tan θ = sin θ, find the value of sin² θ – cos² θ.

Question 11. If cosec θ = 13/12, find the value of $\frac{2sinθ-3cosθ}{4sinθ-9cosθ}$ .

Question 12. If sin θ + cos θ = √2 cos (90^o–θ), find cot θ.

Question 13. If 2sin² θ – cos² θ = 2, then find the value of θ.

Question 14. If √3tan θ – 1 = 0, find the value of sin² θ – cos² θ.