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

• Last Updated : 28 Apr, 2021

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

Solution:

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

Now we know,

=> sin Î¸ = âˆš(1 â€“ cos2 Î¸)

=> sin Î¸ = âˆš(1 â€“ (4/5)2)

=> sin Î¸ = âˆš(1 â€“ (16/25))

=> sin Î¸ = âˆš(9/25)

=> sin Î¸ = 3/5

So, cosec Î¸ = 1/sin Î¸ = 5/3

And tan Î¸ = sin Î¸/cos Î¸ = (3/5)/(4/5) = 3/4

Therefore, cot Î¸ = 1/tan Î¸ = 4/3

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

Question 2. If sin Î¸ = 1/âˆš2, find all other trigonometric ratios of angle Î¸.

Solution:

We are given, sin Î¸ = 1/âˆš2. So, cosec Î¸ =1/sin Î¸ = âˆš2.

Now we know,

=> cos Î¸ = âˆš(1 â€“ sin2 Î¸)

=> cos Î¸ = âˆš(1 â€“ (1/âˆš2)2)

=> cos Î¸ = âˆš(1 â€“ (1/2))

=> cos Î¸ = âˆš(1/2)

=> cos Î¸ = 1/âˆš2

So, sec Î¸ = 1/cos Î¸ = âˆš2

And tan Î¸ = sin Î¸/cos Î¸ = (1/âˆš2)/(1/âˆš2) = 1

Therefore, cot Î¸ = 1/tan Î¸ = 1

If sin Î¸ = 1/âˆš2, value of cosec Î¸, cos Î¸, sec Î¸, tan Î¸ and cot Î¸ are âˆš2, 1/âˆš2, âˆš2, 1 and 1 respectively.

Question 3. If tan Î¸ = 1/âˆš2, find the value of .

Solution:

We are given, tan Î¸ = 1/âˆš2. Now we know,

=> sec Î¸ = âˆš(1 + tan2 Î¸)

=> sec Î¸ = âˆš(1 + (1/âˆš2)2)

=> sec Î¸ = âˆš(1+(1/2))

=> sec Î¸ = âˆš(3/2)

And cot Î¸ = 1/tan Î¸ = âˆš2. Also, we know,

=> cosec Î¸ = âˆš(1 + cot2 Î¸)

=> cosec Î¸ = âˆš(1 + (âˆš2)2)

=> cosec Î¸ = âˆš(1 + 2)

=> cosec Î¸ = âˆš3

So,   =

Therefore, the value of  is .

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

Solution:

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

=> sec Î¸ = âˆš(1 + tan2 Î¸)

=> sec Î¸ = âˆš(1 + (3/4)2)

=> sec Î¸ = âˆš(1+(9/16))

=> sec Î¸ = âˆš(25/16)

=> sec Î¸ = 5/4

And cos Î¸ = 1/sec Î¸ = 4/5.

So,  =

Therefore, the value of  is .

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

Solution:

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

Now we know,

=> cosec Î¸ = âˆš(1 + cot2 Î¸)

=> cosec Î¸ = âˆš(1 + (5/12)2)

=> cosec Î¸ = âˆš(1 + (25/144))

=> cosec Î¸ = âˆš(169/144)

=> cosec Î¸ = 13/12

And sin Î¸ = 1/cosec Î¸ = 12/13.

So,  =

= 25

Therefore, the value of  is 25.

Question 6. If cot Î¸ = 1/âˆš3, find the value of .

Solution:

We are given, cot Î¸ = 1/âˆš3. Now we know,

=> cosec Î¸ = âˆš(1 + cot2 Î¸)

=> cosec Î¸ = âˆš(1 + (1/âˆš3)2)

=> cosec Î¸ = âˆš(1 + (1/3))

=> cosec Î¸ = âˆš(4/3)

=> cosec Î¸ = 2/âˆš3

And sin Î¸ = 1/cosec Î¸ = âˆš3/2. Also, we know,

=> cos Î¸ = âˆš(1 â€“ sin2 Î¸)

=> cos Î¸ = âˆš(1 â€“ (âˆš3/2)2)

=> cos Î¸ = âˆš(1 â€“ (3/4))

=> cos Î¸ = âˆš(1/4)

=> cos Î¸ = 1/2

So,  =

Therefore, the value of  is .

Question 7. If cosec A = âˆš2, find the value of .

Solution:

We are given, cosec A = âˆš2. So, sin A = 1/cosec A = 1/âˆš2.

Now we know,

=> cos A = âˆš(1 â€“ sin2 A)

=> cos A = âˆš(1 â€“ (1/âˆš2)2)

=> cos A = âˆš(1 â€“ (1/2))

=> cos A = âˆš(1/2)

=> cos A = 1/âˆš2

Hence, tan A = sin A/cos A = (1/âˆš2)/(1/âˆš2) = 1. And cot A = 1/tan A = 1.

So,  =

= 2

Therefore, the value of  is 2.

Question 8. If cot Î¸ = âˆš3, find the value of .

Solution:

We are given cot Î¸ = âˆš3. And tan Î¸ = 1/cot Î¸ = 1/âˆš3.

Now we know,

=> cosec Î¸ = âˆš(1 + cot2 Î¸)

=> cosec Î¸ = âˆš(1 + (âˆš3)2)

=> cosec Î¸ = âˆš4

=> cosec Î¸ = 2

Also, we know,

=> sec Î¸ = âˆš(1 + tan2 Î¸)

=> sec Î¸ = âˆš(1 + (1/âˆš3)2)

=> sec Î¸ = âˆš(1+(1/3))

=> sec Î¸ = âˆš(4/3)

=> sec Î¸ = 2/âˆš3

So,  =

Therefore, the value of  is .

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

Solution:

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

=> sin Î¸ = âˆš(1 â€“ cos2 Î¸)

=> sin Î¸ = âˆš(1 â€“ (1/3)2)

=> sin Î¸ = âˆš(1 â€“ (1/9))

=> sin Î¸ = âˆš(8/9)

=> sin Î¸ = 2âˆš2/3

Hence, tan Î¸ = sin Î¸/cos Î¸ = (2âˆš2/3)/(1/3) = 2âˆš2

So,  =

= 10

Therefore, the value of  is 10.

Question 10. If âˆš3 tan Î¸ = sin Î¸, find the value of sin2 Î¸ â€“ cos2 Î¸.

Solution:

We are given, âˆš3 tan Î¸ = sin Î¸

=> âˆš3 (sin Î¸/cos Î¸) = sin Î¸

=> cos Î¸ = 1/âˆš3

Now we know,

=> sin Î¸ = âˆš(1 â€“ cos2 Î¸)

=> sin Î¸ = âˆš(1 â€“ (1/âˆš3)2)

=> sin Î¸ = âˆš(1 â€“ (1/3))

=> sin Î¸ = âˆš(2/3)

So, sin2 Î¸ â€“ cos2 Î¸ = âˆš(2/3)2 â€“ (1/âˆš3)2

= 2/3 â€“ 1/3

= 1/3

Therefore, the value of sin2 Î¸ â€“ cos2 Î¸ is 1/3.

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

Solution:

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

Now we know,

=> cos Î¸ = âˆš(1 â€“ sin2 Î¸)

=> cos Î¸ = âˆš(1 â€“ (12/13)2)

=> cos Î¸ = âˆš(1 â€“ (144/169))

=> cos Î¸ = âˆš(25/169)

=> cos Î¸ = 5/13

So,  =

= 3

Therefore, the value of  is 3.

Question 12. If sin Î¸ + cos Î¸ = âˆš2 cos (90oâ€“Î¸), find cot Î¸.

Solution:

We are given,

=> sin Î¸ + cos Î¸ = âˆš2 cos (90oâ€“Î¸)

=> sin Î¸ + cos Î¸ = âˆš2 sin Î¸

=> cos Î¸ = (âˆš2â€“1)sin Î¸

=> cos Î¸/sin Î¸ = âˆš2â€“1

=> cot Î¸ = âˆš2â€“1

Therefore, value of cot Î¸ is âˆš2â€“1.

Question 13. If 2sin2 Î¸ â€“ cos2 Î¸ = 2, then find the value of Î¸.

Solution:

We have,

=> 2sin2 Î¸ â€“ cos2 Î¸ = 2

=> 2sin2 Î¸ â€“ (1 â€“ sin2 Î¸) = 2

=> 2sin2 Î¸ â€“ 1 + sin2 Î¸ = 2

=> 3sin2 Î¸ = 3

=> sin2 Î¸ = 1

=> sin Î¸ = 1

=> Î¸ = 90o

Therefore, the value of Î¸ is 90o.

Question 14. If âˆš3tan Î¸ â€“ 1 = 0, find the value of sin2 Î¸ â€“ cos2 Î¸.

Solution:

We are given,

=> âˆš3tan Î¸ â€“ 1 = 0

=> tan Î¸ = 1/âˆš3

=> tan Î¸ = tan 30o

=> Î¸ = 30o

So, sin2 Î¸ â€“ cos2 Î¸ = sin2 30o â€“ cos2 30o

= (1/2)2 â€“ (âˆš3/2)2

= (1/4) â€“ (3/4)

= â€“2/4

= â€“1/2

Therefore, the value of sin2 Î¸ â€“ cos2 Î¸ is â€“1/2.

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