Question

if a b and c are interior angles of a triangle abc then show that cosec 2(b+c/2) -tan2 a/2 =1

Answer 1

We know that Sum of interior angles in a triangle equals to 180°,
=> a + b + c = 180°,
Simplifying,

=> b+c = 180° - a,
=> (b+c)/2 = (90° - a/2),

We know these formulas :
(1) Cosec ( 90° - A) = Sec A,
(2) Sec² A - Tan² = 1,

Given that,
Required To Prove : Cosec² ((b+c)/2) - Tan² (a/2) = 1,
Simplifying LHS to prove,

=> We can write (b+c)/2 as 90-a/2 Remember?
=> Cosec² (90-a/2) - Tan² (a/2),
=> We know Cosec A = Sec(90-A) and vice versa, Remember?
=> Sec² (a/2) - Tan² (a/2),
We know Sec² A - Tan² A = 1, Remember?
=> Sec² (a/2) - Tan² (a/2) = 1,

We have got the LHS as 1, by simplifying it, And what we needed to prove was, LHS = 1, and we have done it !

Therefore, HENCE Proved it !

Hope you understand, Have a Great Day ! Merry Christmas !
Thanking you, Bunti 360 !

Answer 2

A + B + C = π 
(B + C)/2 = (π/2) - (A/2) 

csc²{(B+C)/2} - tan²{A/2} 
= [1/sin{(B+C)/2}]² - [sin²{A/2} / cos²{A/2}] 
= [1/sin{(π/2) - (A/2)}]² - [sin²{A/2} / cos²{A/2}] 
= [1/cos²{A/2}] - [sin²{A/2} / cos²{A/2}] 
= [1 - sin²{A/2}] / cos²{A/2} 
= cos²{A/2} / cos²{A/2} 
= 1