Question

if A B C D are angles of a quadrilateral prove that tan (A+ B/2) + tan (C+ D/2) = 0 ​

Answer 1

Answer:

Step-by-step explanation:

Given :

To prove :

If A , B , C & D are angles of a Quadrilateral,

Then,

$tan(\frac{A+B}{2}) + tan (\frac{C+D}{2}) = 0$

Solution :

We know that,

Tan (180° - θ) = -Tanθ = Tan(π - θ)

Sum of the internal angles on a Quadrilateral = 360° = 2π radians,.

⇒ ∠A + ∠B + ∠C + ∠D = 360° = 2π

⇒ ∠A + ∠B = 360° - (∠C + ∠D) = 2π - (∠C + ∠D)

Dividing both the sides, by 2,

We get,

⇒ $\frac{A+B}{2} = 180 - \frac{C+D}{2}$

In tangent (Tan) function,.

Tan (180° - θ) = -Tanθ

⇒ $tan(\frac{A+B}{2}) = tan (180 - \frac{C+D}{2})$

⇒ $tan(\frac{A+B}{2}) = -tan (\frac{C+D}{2})$

⇒ $tan(\frac{A+B}{2}) + tan (\frac{C+D}{2}) = 0$

Hence, Proved,.

Answer 2

Answer:

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