prove that whole root of 1 -cos2tita/ 1+cos2tita root = tan tita
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
Solution:
Taking the L.H.s,
(1-cos2 A) cosec2 A
(sin2 A) cosec2 A [" sin2 A + cos2 A =11
- 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 +
Taking,
L.H.S = (1+ cot2 A) sin2 A
= cosec2 A sin2 A
= (cosec A sin A)2
(1/sin A) * sin Aj2
= (1)2
=1
R.H.S
-Hence Proved
3. tan2 e cos2 9 =1-cos2
Solution:
We know that,
Sinz et coS2 61
Taking,
L.H.S =tan2 6 cos2 6
(tan& x Cos 8)2
(sin ).
= sin2 e
1-cos2 9
R.H.S
-Hence Proved
4. cosec 9 v(1- cos2 8) =1
Solution:
Using identity,
Sin2 e coS2 6 =19 Sin2 6=1-co52 6
Taking L.H.S.
L.H.S = Cosec e {1- cos2 68)
COsec 6 { sin2 e)
cosec x sin
=1
= R.H.S
- Hence Proved
5. (sec2 6- 1(cosec26-1)=1
Solution
Using identities,
(sec2 - tan2 6) = 1 and (cosec2 6 - cot2 e) =
1
We have,
LH.S = (sec2 - 1(cosec20 1)
= tan26 x cot26
(tan x cot 6)2
(tan e x 1tan 9)2
= 12
= 1
R.H.S
Hence Proved
6. tan +1/ tan 6 = sec 6 cosec
Solution
We have,
L.H.S = tan +1/ tan e
= (tan2 + 1/ tan
sec2 9/ tan 8 [ sec2 9- tan2 e =11
(1cos2 6) x V (sin e/cos 6) [: tan 6 = Sin 6
cos 6
= COs 6/ (sin 6 x cos2 8)
1/ cos 9 x 1/ sin 9
sec 8 x cosec
sec 6 cosec 6
R.H.S
-- Hence proved.....