Class 10 RD Sharma Solutions Chapter 6 Trigonometric Identities Exercise 6.2 (original) (raw)

Last Updated : 23 Jul, 2025

In Chapter 6 of RD Sharma's Class 10 Mathematics textbook, we explore Trigonometric Identities a fundamental topic in trigonometry. Exercise 6.2 focuses on applying various trigonometric identities to solve problems and simplify expressions. Understanding these identities is crucial for solving complex trigonometric equations and proofs which are essential skills in mathematics.

Geometric Progressions

A geometric progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous term by the constant called the common ratio. For example: in the sequence 2, 6, 18, 54 the common ratio is 3. The general formula for the n-th term of a GP is given by an=a⋅r(n−1) wherea is the first term and r is the common ratio.

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 sin 2 θ – cos 2 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 (90o–θ), find cot θ.

**Solution:

We are given,

=> sin θ + cos θ = √2 cos (90o–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 2sin 2θ – cos2 θ = 2, then find the value of θ.

**Solution:

We have,

=> 2sin2 θ – cos2 θ = 2

=> 2sin2 θ – (1 – sin2 θ) = 2

=> 2sin2 θ – 1 + sin2 θ = 2

=> 3sin2 θ = 3

=> sin2 θ = 1

=> sin θ = 1

=> θ = 90o

**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 30o

=> θ = 30o

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

= (1/2)2 – (√3/2)2

= (1/4) – (3/4)

= –2/4

= –1/2

**Therefore, the value of sin 2 θ – cos 2 θ is –1/2.

**Read More:

• Trigonometric Identities
• Trigonometry in Maths: Table, Formulas, Identities and Ratios

Conclusion

Exercise 6.2 in Chapter 6 of RD Sharma's textbook helps reinforce the application of the trigonometric identities in the various problems. The Mastering these identities enables students to the simplify complex trigonometric expressions and solve equations efficiently. A solid understanding of trigonometric identities is essential for the tackling more advanced mathematical concepts.