A B and C are interior angles of triangle ABC then show that tan((A B)/2=COT C/2
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In a ΔABC, ∠A + ∠B + ∠C = 180° [Angle sum property]
⇒ ∠A + ∠B = 180° - ∠C
⇒ (∠A + ∠B) / 2 = (180° - ∠C) / 2
⇒ (∠A + ∠B) / 2 = (180° - ∠C) / 2
⇒ (∠A + ∠B) / 2 = (90° - ∠C/ 2)
⇒ sin (∠A + ∠B) / 2 = sin (90° - ∠C/ 2)
⇒ sin (∠A + ∠B) / 2 = cos ∠C/ 2. [sin (90 - θ) = cos θ)]
Therefore, sin (∠A + ∠B) / 2 in terms of angle c is cos ∠C/ 2.
In a ΔABC, ∠A + ∠B + ∠C = 180° [Angle sum property]
⇒ ∠A + ∠B = 180° - ∠C
⇒ (∠A + ∠B) / 2 = (180° - ∠C) / 2
⇒ (∠A + ∠B) / 2 = (180° - ∠C) / 2
⇒ (∠A + ∠B) / 2 = (90° - ∠C/ 2)
⇒ sin (∠A + ∠B) / 2 = sin (90° - ∠C/ 2)
⇒ sin (∠A + ∠B) / 2 = cos ∠C/ 2. [sin (90 - θ) = cos θ)]
Therefore, sin (∠A + ∠B) / 2 in terms of angle c is cos ∠C/ 2.
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