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VOOZH | about |
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.
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.
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.
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.
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, =
=
=
Therefore, the value of is .
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, =
=
=
Therefore, the value of is .
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, =
=
= 25
Therefore, the value of is 25.
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, =
=
=
=
Therefore, the value of is .
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, =
=
= 2
Therefore, the value of is 2.
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, =
=
=
Therefore, the value ofis .
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, =
=
=
= 10
Therefore, the value of is 10.
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.
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, =
=
=
= 3
Therefore, the value of is 3.
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.
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 90o.
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 sin2 θ – cos2 θ is –1/2.
Read More:
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.
