Question

If A, B, C are the interior angles of a triangle ABC, prove that
(i)$tan(\frac{C+A}{2})=cot\frac{B}{2}$
(ii)$sin(\frac{B+C}{2})=cos\frac{A}{2}$

Answer 1

SOLUTION :  

(i) Given :  If A, B, C are the interior angles of a triangle ABC.  

prove : tan(C + A)/2 = cot B/2

In ∆ ABC,

A + B + C = 180°

[Sum of all angles of a ∆ is equal to 180°]

(C + A) = 180° - B

(C + A) /2 = 180°/2 − B/2

[On Dividing both Sides by 2]

(C + A) /2 = 90° − B/2

On taking tan both sides,

tan (C + A) /2 = tan(90° − B/2)

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

[tan (90° - θ) = cot θ]

Hence proved

 

(ii) Given :If A, B, C are the interior angles of a triangle ABC.

prove : sin(B + C)/2 = cos A/2

In ∆ ABC,

A + B + C=180°

[Sum of the angles of a ∆ is 180°]

B + C = 180° - A

(B+ C)/2 = 180° /2 − A/2

[On Dividing both Sides by 2]

(B+ C)/2 = 90° /2 −A/2

On taking sin both sides,

sin (B+C)/2 = sin (90° - A)/2

sin(B+C)/2 = cosA/2

[sin (90 - θ) = cos θ]

Hence proved

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Answer 2

QUESTION:-

If A, B, C are the interior angles of a triangle ABC, prove that
(i)$tan(\frac{C+A}{2})=cot\frac{B}{2}$
(ii)$sin(\frac{B+C}{2})=cos\frac{A}{2}$

ANSWER:- ↖️⬆️⬆️⬆️⬆️⬆️↗️