if cosec tetha 1 = sec theta 2 then show that theta 1 + theta 2 =90 degrees
Answers
1. (1 – cos2 A) cosec2 A = 1
Solution:
Taking the L.H.S,
(1 – cos2 A) cosec2 A
= (sin2 A) cosec2 A [∵ sin2 A + cos2 A = 1 ⇒1 – sin2 A = cos2 A]
= 12
= 1 = R.H.S
– Hence Proved
2. (1 + cot2 A) sin2 A = 1
Solution:
By using the identity,
cosec2 A – cot2 A = 1 ⇒ cosec2 A = cot2 A + 1
Taking,
L.H.S = (1 + cot2 A) sin2 A
= cosec2 A sin2 A
= (cosec A sin A)2
= ((1/sin A) × sin A)2
= (1)2
= 1
= R.H.S
– Hence Proved
3. tan2 θ cos2 θ = 1 − cos2 θ
Solution:
We know that,
sin2 θ + cos2 θ = 1
Taking,
L.H.S = tan2 θ cos2 θ
= (tan θ × cos θ)2
= (sin θ)2
= sin2 θ
= 1 – cos2 θ
= R.H.S
– Hence Proved
4. cosec θ √(1 – cos2 θ) = 1
Solution:
Using identity,
sin2 θ + cos2 θ = 1 ⇒ sin2 θ = 1 – cos2 θ
Taking L.H.S,
L.H.S = cosec θ √(1 – cos2 θ)
= cosec θ √( sin2 θ)
= cosec θ x sin θ
= 1
= R.H.S
– Hence Proved
5. (sec2 θ − 1)(cosec2 θ − 1) = 1
Solution:
Using identities,
(sec2 θ − tan2 θ) = 1 and (cosec2 θ − cot2 θ) = 1
We have,
L.H.S = (sec2 θ – 1)(cosec2θ – 1)
= tan2θ × cot2θ
= (tan θ × cot θ)2
= (tan θ × 1/tan θ)2
= 12
= 1
= R.H.S
– Hence Proved
6. tan θ + 1/ tan θ = sec θ cosec θ
Solution:
We have,
L.H.S = tan θ + 1/ tan θ
= (tan2 θ + 1)/ tan θ
= sec2 θ / tan θ [∵ sec2 θ − tan2 θ = 1]
= (1/cos2 θ) x 1/ (sin θ/cos θ) [∵ tan θ = sin θ / cos θ]
= cos θ/ (sin θ x cos2 θ)
= 1/ cos θ x 1/ sin θ
= sec θ x cosec θ
= sec θ cosec θ
= R.H.S
– Hence Proved
7. cos θ/ (1 – sin θ) = (1 + sin θ)/ cos θ
Solution:
We know that,
sin2 θ + cos2 θ = 1
So, by multiplying both the numerator and the denominator by (1+ sin θ), we get
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 1
L.H.S =
= R.H.S
– Hence Proved
8. cos θ/ (1 + sin θ) = (1 – sin θ)/ cos θ
Solution:
We know that,
sin2 θ + cos2 θ = 1
So, by multiplying both the numerator and the denominator by (1- sin θ), we get
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 2
L.H.S =
= R.H.S
– Hence Proved
9. cos2 θ + 1/(1 + cot2 θ) = 1
Solution:
We already know that,
cosec2 θ − cot2 θ = 1 and sin2 θ + cos2 θ = 1
Taking L.H.S,
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 3
= cos2 A + sin2 A
= 1
= R.H.S
– Hence Proved
10. sin2 A + 1/(1 + tan 2 A) = 1
Solution:
We already know that,
sec2 θ − tan2 θ = 1 and sin2 θ + cos2 θ = 1
Taking L.H.S,
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 4
= sin2 A + cos2 A
= 1
= R.H.S
– Hence Proved
11.
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 5
Solution:
We know that, sin2 θ + cos2 θ = 1
Taking the L.H.S,
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 6
= R.H.S
– Hence Proved
12. 1 – cos θ/ sin θ = sin θ/ 1 + cos θ
Solution:
We know that,
sin2 θ + cos2 θ = 1
So, by multiplying both the numerator and the denominator by (1+ cos θ), we get
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 7
= R.H.S
– Hence Proved
13. sin θ/ (1 – cos θ) = cosec θ + cot θ
Solution:
Taking L.H.S,
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 8
= cosec θ + cot θ
= R.H.S
– Hence Proved
14. (1 – sin θ) / (1 + sin θ) = (sec θ – tan θ)2
Solution:
Taking the L.H.S,
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 9
= (sec θ – tan θ)2
= R.H.S
– Hence Proved
15. R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 10
Solution:
Taking L.H.S,
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 11
= cot θ
= R.H.S
– Hence Proved
16. tan2 θ − sin2 θ = tan2 θ sin2 θ
Solution:
Taking L.H.S,
L.H.S = tan2 θ − sin2 θ
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 12
= tan2 θ sin2 θ
= R.H.S
– Hence Proved
17. (cosec θ + sin θ)(cosec θ – sin θ) = cot2θ + cos2θ
Solution:
Taking L.H.S = (cosec θ + sin θ)(cosec θ – sin θ)
On multiplying we get,
= cosec2 θ – sin2 θ
= (1 + cot2 θ) – (1 – cos2 θ) [Using cosec2 θ − cot2 θ = 1 and sin2 θ + cos2 θ = 1]
= 1 + cot2 θ – 1 + cos2 θ
= cot2 θ + cos2 θ
= R.H.S
– Hence Proved
18. (sec θ + cos θ) (sec θ – cos θ) = tan2 θ + sin2 θ
Solution:
Taking L.H.S = (sec θ + cos θ)(sec θ – cos θ)
On multiplying we get,
= sec2 θ – sin2 θ
= (1 + tan2 θ) – (1 – sin2 θ) [Using sec2 θ − tan2 θ = 1 and sin2 θ + cos2 θ = 1]
= 1 + tan2 θ – 1 + sin2 θ
= tan 2 θ + sin 2 θ
= R.H.S
– Hence Proved
19. sec A(1- sin A) (sec A + tan A) = 1
Solution:
Taking L.H.S = sec A(1 – sin A)(sec A + tan A)
Substituting sec A = 1/cos A and tan A =sin A/cos A in the above we have,
L.H.S = 1/cos A (1 – sin A)(1/cos A + sin A/cos A)
= 1 – sin2 A / cos2 A [After taking L.C.M]
= cos2 A / cos2 A [∵ 1 – sin2 A = cos2 A]
= 1
= R.H.S
– Hence Proved
20. (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1
Solution:
Taking L.H.S = (cosec A – sin A)(sec A – cos A)(tan A + cot A)