Question

If A, B, C are the angles of ∆ABC then prove that Cosec² [(B+C)/2]–tan²A/2 =1

Answer 1

Trigonometry is the study of the relationship between the sides and angles of a triangle.

Two angles are said to be complementary of their sum is equal to 90° .
θ & (90° - θ) are complementary angles.

An equation involving trigonometry ratios of an angle is called is called a trigonometric  identity, if it is true for all values of the angles involved. For any acute angle θ, we have 3 identities.

i) sin² θ + cos² θ = 1 ,ii) 1 + tan² θ = sec² θ , iii) cot² θ +1 = cosec² θ.

SOLUTION:
We know that the sum of the angles of triangle is 180°

In ∆ABC ,
A + B + C = 180°
B + C = 180° - A
(B + C )/2 = (180° - A)/2

[Dividing by 2 on both sides]

(B + C )/2 = 90° - A/2……….(1)

LHS = Cosec² [(B+C)/2] - tan²A/2
= Cosec² [90° - A/2] – tan²A/2

[ From eq 1]

= sec² A/2 - tan²A/2 = 1

[Cosec θ = Sec(90-θ)] , Sec² θ - tan² θ = 1]
Cosec² [(B+C)/2] - tan²A/2  = 1

LHS = RHS

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