**If 3θ is an acute angle, solve the following equations for θ:**

(i) (cosec 3θ – 2) (cot 2θ – 1) = 0

(ii) (tan θ – 1) (cosec 3θ – 1) = 0

**Answer :**

**(i)** (cosec 3θ – 2) (cot 2θ – 1) = 0

cosec 3θ – 2 or cot 2θ – 1 = 0

⇒ cosec 3θ = 2 or cot 2θ = 1

So,

cosec 3θ = cosec 3θ° or cot 2θ =cot 45°

⇒ 3θ = 30° or 2θ = 45°

Thus,

θ = 30° or 45°.

**(ii)** (tan θ – 1) (cosec 3θ – 1) = 0

tan θ – 1 = 0 or cosec 3θ – 1 = 0

⇒ tan θ = 1 or cosec 3θ = 1

So,

tan θ = tan45° or cosec 3 θ = cosec 90°

⇒ θ = 45° or 3θ = 90° i.e. θ = 30°

Thus,

θ = 45° or 30°.

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