Q12 of 268 Page 4

If tan 3 θ =cot (θ +18^o), where 3 θ and θ +18^o are acute angles, find the value of θ.

tan 3θ = cot (θ + 18°) …(i)

We know that

tan θ = cot (90° - θ)

So, Eq. (i) become

Cot (90° - 3θ) = cot (θ + 18°)

On Equating both the sides, we get

90° - 3θ = θ + 18°

⇒ -3θ - θ = 18° -90°

⇒ -4θ = -72°

⇒ θ = 18°

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If tan 3 θ =cot (θ +18o), where 3 θ and θ +18o are acute angles, find the value of θ.