Question

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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 !

Answer 2

Brainly.in

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aman7863
Secondary SchoolMath 5+3 pts


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aman7863
aman7863
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hemantvats17
Hemantvats17 Genius
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 !