Question

in triangle ABC;

The value of

$\sf\tan( \frac{B + C}{2} ) \tan( \frac{A}{2} ) + \tan( \frac{A + C}{2} ) \tan( \frac{B}{2} ) + \tan(\frac{B + A}{2} ) \tan( \frac{C}{2} )$

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Answer 1

We know that, the sum of all interior angles of a triangle is 180°.

Here in triangle ABC, we can say that angles A + B + C = 180°.

On dividing both sides by 2, we get,

(A + B + C)/2 = 180° / 2

=> A/2 + B/2 + C/2 = 90°

From this, we derive,

→ (B + C)/2 = 90° - (A/2)

→ (A + C)/2 = 90° - (B/2)

→ (B + A)/2 = 90° - (C/2)

So,

→ tan ((B + C)/2) = tan (90° - (A/2)) = cot (A/2)

→ tan ((A + C)/2) = tan (90° - (B/2)) = cot (B/2)

→ tan ((B + A)/2) = tan (90° - (C/2)) = cot (C/2)

Now,

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

=> cot (A/2) tan (A/2) + cot (B/2) tan (B/2) + cot (C/2) tan (C/2)

And, since tangent and cotangent values are reciprocals to each other, we get,

=> 1 + 1 + 1 = 3

Hence 3 is the answer.

Answer 2

Answer:

3is answer

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