Question

question 19) (1) and (2)

Answer 1

A , B, And C are the interior Angles of a ∆ABC.

To prove : (1) Cos ( A + B /2 ) = Sin C/2

We know that the sum of the angles of a triangle is 180°.

Therefore,

=> A + B + C = 180

=> A + B = ( 180 - C )

=> A + B/2 = 180 - C /2

=> A + B/2 = 90 - C/2

=> Cos ( A + B/2 ) = Cos ( 90 - C/2 )

=> Cos ( A + B/2 ) = Sin C /2 [ Since Cos(90- Theta) = Sin theta ]

Hence

Cos ( A + B/2 ) = Sin C/2 [ Proved ]

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(2) tan C + A/2 = Cot B/2

=> A + B + C = 180

=> C + A = ( 180 - B )

=> ( C + A /2 ) = ( 180 - B /2 )

=> ( C + A /2 ) = 90 - B /2

=> Tan ( C + A /2 ) = Tan ( 90 - B/2 )

=> Tan ( C + A/2 ) = Cot B/2 [ Since Tan ( 90 - Theta ) = Cot ¢ ]

Hence,

Tan ( C + A /2 ) = Cot B/2 [ Proved ]



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