In a triangle ABC - cos - A - B - 2- is equal toA- cos C - 2B- sin C - 2C- Sec C-2D- tan C - 2

The correct option is B sin C/2 nbsp; ...

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Calculating Heights and Distances

In a triangle...

Question

In a triangle ABC, cos[(A+B)/2)] is equal to

cos C/2

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sin C/2

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Sec C/2

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tan C/2

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Solution

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Similar questions

△ A B C

s i n (\frac{A + B}{2})

In triangle ABC, $1 - t a n \frac{B}{2} . t a n \frac{C}{2}$ is equal to

Δ A B C, \frac{a + b}{c} =

△ A B C

s i n (\frac{A + B}{2}) =

△ A B C,

A, B

C

cos A + cos B + cos C

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