Question

If A,B,C are interior angles of a triangle ,then prove that. Tan[A+B/2]=>Cot[C/2]​

Answer 1

Answer:

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In a triangle, sum of interior angles=180⁰

Α+B+C=180⁰

Α+B. =180⁰-C

(A+B)/2=(90⁰-C/2)

Taking tangent

tan(A+B)/2=tan(90⁰-C/2)

tan(A+B)/2=Cot(C/2)

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Answer 2

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⭐${\blue{Explanation}}$⭐

✅✅➡️✨✔✔A,B,C are interior angles of a triangle

So LA + LB + LC = 180⁰

LA + LB + LC = 180⁰

LA + LB = 180⁰ - LC

Tan[a+b/2] can be written as cot[90⁰-(a+b/2)] because A, B, C are acute angles of triangle.

so cot[90⁰-(a+b/2)]can be written as

cot[90⁰-(180⁰-C/2)] as LA + LB = 180⁰ -LC

so, it becomes cot[90⁰-(180⁰-LC/2)]

cot[90⁰-(90-C/2)]

cot[90⁰-90⁰+C/2]

cot[C/2] = RHS✨✨

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