please solve this ques please
Answers
1. 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)
2 and 3.
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 !
By angle sum property :
A + B + C = 180°
( i )
B + C = 180 - A
⇒ ( B + C )/2 = 90 - A/2
Take sin on both sides :
sin ( B + C ) = sin ( 90 - A/2 )
Note that sin ( 90 - x ) = cos x
⇒ sin ( B + C )/2 = cos A/2
Hence proved !
( ii )
A + B = 180 - C
⇒ ( A + B )/2 = 90 - C/2
Take cosec on both sides :
⇒ cosec ( A + B )/2 = sec( 90 - C/2 )
We know that sec ( 90 - x ) = cosec x
⇒ cosec ( A + B )/2 = cosec C/2
( iii )
A + C = 180 - B
⇒ ( A + C )/2 = 90 - B/2
Take sec on both sides :
⇒ sec ( A + C )/2 = sec ( 90 - B/2 )
sec ( 90 - x ) = cosec x