Question

A, B, C are the interior angles of a triangle. Show that:

1. cosec (A+B/2) = sec (C/2).​

Answer 1

Answer:

Step-by-step explanation: for 2 marks

A+B+C=180 (angle sum property)

A+B=180-C

LHS=cosec(A+B)/2

=cosec(180-C)/2

=cosec 90-C/2

=sec C/2

HENCE PROVED

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Answer 2

Step-by-step explanation:

In ∆ABC ,

Sum of all interior angles = 180°

=> <A + <B + <C = 180°

=> <A + <B = 180° - <C

=> (<A + <B)/2 = (180° - <C)/2

=> (<A + <B)/2 = 90° - <C/2

Now,

cosec((A + B)/2) = cosec( 90° - C/2) = sec(C/2)

[ cosec(90° - P) = sec(P)]

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