Question

If A,B,C are angle of a triangle then find i) sin(a+b)/2
ii) cosec(b+c)/2​

Answer 1

Question :

If A, B, C are angles of ∆ABC the prove that;

• i) sin [(A+B)/(2)] = cos (C)/(2)
• ii) cosec [(B+C)/(2)] = sec (A)/(2)

ANSWER

Given : -

A, B, C are angles of ∆ABC

Required to prove : -

• i) sin [(A+B)/(2)] = cos (C)/(2)
• ii) cosec [(B+C)/(2)] = sec (A)/(2)

Proof : -

A, B, C are angles of ∆ABC

We need to prove that;

• i) sin [(A+B)/(2)] = cos (C)/(2)
• ii) cosec [(B+C)/(2)] = sec (A)/(2)

So,

Let's prove the 1st bit (i)

We know that,

In a ∆ABC,

∠A+∠B+∠C = 180° ( Angle Sum Property)

Dividing by 2 on both sides

(∠A+∠B+∠C)/(2) = (180°)/(2)

(∠A+∠B+∠C)/(2) = 90°

(∠A+∠B)/(2)+(∠C)/(2) = 90°

Transposing (∠C)/(2) on the right side

This implies;

(∠A+∠B)/(2) = 90°-(∠C)/(2)

Taking sin function on both sides

sin [(∠A+∠B)/(2)] = sin [90°-(∠C)/(2)]

Since,

• sin (90°-A) = cos A

sin [(∠A+∠B)/(2)] = cos (∠C)/(2)

Hence Proved !

Now,

(ii)

We know that,

In a ∆ABC,

∠A+∠B+∠C = 180° ( Angle Sum Property)

Dividing by 2 on both sides

(∠A+∠B+∠C)/(2) = (180°)/(2)

(∠A+∠B+∠C)/(2) = 90°

(∠A)/(2)+(We know that,

In a ∆ABC,

∠A+∠B+∠C = 180° ( Angle Sum Property)

Dividing by 2 on both sides

(∠A+∠B+∠C)/(2) = (180°)/(2)

(∠A+∠B+∠C)/(2) = 90°

(∠A)/(2)+(∠B+∠C)/(2) = 90°

Transposing (∠A)/(2) on the right side

This implies;

(∠B+∠C)/(2) = 90°-(∠A)/(2)

Taking cosec function on both sides

cosec [(∠B+∠C)/(2)] = sec [90°-(∠A)/(2)]

Since,

• cosec (90°-A) = sec A

cosec [(∠B+∠C)/(2)] = sec (∠A)/(2)

Hence Proved !

Answer 2

Step-by-step explanation:

1 Question : -

= sin(A+B/2)

A + B + C=180

A + B = 180 - C

A + B /2 = 180 - C /2

= 180 /2 - C /2

=(90 - C /2)

sin(90 - C /2)

sin(90 - A) = cos A

= cos C /2

2 Question : -

A + B + C = 180

A + B = 180 - C

LHS = cosec (A + B)/2

= cosec(180 - C)/2

= cosec 90 - C/2

= sec C /2