Solve
(i) sin9 θ = sin θ
(ii) cot θ + tan θ = 2cosec θ
(iii) sin2 θ = cos3 θ
(iv) cot θ = tan8 θ
(v) cot θ – tan θ = 2.
(vi) cosec θ = cot θ + 
(vii) tan2 θ tan θ = 1
(viii) tan θ + tan2 θ +
tan θ tan2 θ =
.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i)sin 9 θ = sin θ ⇒ sin 9 θ – sin θ = 0 ⇒ 2 cos 5 θ sin 4 θ = 0 ⇒ cos 5 θ = 0 or
sin 4 θ = 0 ⇒ 5 θ = (2n + 1)
or 4 θ = m π⇒ θ = (2n + 1)
or θ =
.
(ii)cot θ + tan θ = 2cosec θ ⇒
+
= 
⇒
=
⇒ 1 = 2cos θ⇒
α
(iii)sin 2 θ = cos 3 θ ⇒ cos
= cos 3 θ ⇒
– 2 θ = 2n π ± 3 θ ⇒
– 2 θ ± 3 θ = 2n π
⇒ θ = 2n π –
,
⇒θ = 2n π –
,

(iv)cot θ = tan8 θ ⇒ tan8 θ = tan
⇒
⇒ 
(v)cot θ – tan θ = 2
⇒ 2 cot 2 θ = 2 ⇒ 2 θ = n π +
⇒ θ =
+
= 
(vi)cosec θ = cot θ + 
⇒
⇒
⇒
= 
⇒
= 0 or tan
=
⇒
or 
⇒
or 
But for
, cosec θ is not defined ∴ 
(vii) tan 2 θ tan θ = 1 ⇒ sin 2 θ sin θ = cos 2 θ cos θ
0 = cos 3 θ ⇒ 3 θ = (2n + 1)
⇒ θ = (2n + 1)
.
(viii) tan θ + tan 2 θ +
tan θ tan 2 θ =
⇒ tan θ + tan 2 θ =
(1 – tan θ tan 2 θ )
=
⇒ tan 3 θ = tan
⇒ 3 θ = n π +
⇒ θ =
+ 
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