Question

If A,B and C are the interior angles of a triangle ABC, show that:

(i) cos A+B/2=Sin/2
(ii) tan C+A/2=cot B/2

Answer 1

$\huge{\underline{\underline{\mathfrak{Answer}}}}$

(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°]

Sin[(B+C)/2]

Since A+B+C=180 for interior angles of

triangle ABC.

then B+C=180-A.

NOW Sin [(180-A)/2]

=Sin[90-(A/2)] since

Sin(90-A)=CosA

=Cos(A/2)