Question

in triangle ABC prove that cosec (a+c/2)=sec(b/2)​

Answer 1

Answer:

Step-by-step explanation:

Given,

ABC is a triangle.

To Prove :-

$cosec\left(\dfrac{A+C}{2}\right)=sec\left(\dfrac{B}{2}\right)$

Formula Required :-

1) Sum of all angles in a triangle = 180°

2) cosec(90° - B/2) = sec(B/2)

Solution :-

ABC is a triangle :-

→ A + B + C = 180°

[ ∴ Sum of all angles in a triangle = 180° ]

A + C = 180° - B

[Let it be equation - 1]

Dividing equation - 1 by '2' :-

$\dfrac{A+C}{2}=\dfrac{180^{\circ}-B}{2}$

Applying 'cosec' on both sides :-

$cosec\left(\dfrac{A+C}{2}\right)=cosec\left(\dfrac{180^{\circ}}{2}-\dfrac{B}{2}\right)$

$cosec\left(\dfrac{A+C}{2}\right)=cosec\left(90^{\circ}-\dfrac{B}{2}\right)$

$cosec\left(\dfrac{A+C}{2}\right)=sec\left(\dfrac{B}{2}\right)$

$\left[\therefore cosec\left(90^{\circ}-\dfrac{B}{2}\right)=sec\left(\dfrac{B}{2}\right)\right]$

Hence Proved that L.H.S = R.H.S.