If $A, B, C$ are the interior angles of a triangle $A B C$ , prove that : $tan \frac{B + C}{2} = cot \frac{A}{2}$

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Solution

If $A, B, C$ are interior angles of a triangle, then

$A + B + C = 180^{\circ}$

$B + C = 180^{\circ} - A$ ........... (1)

Now,

$tan (\frac{B + C}{2})$

$= tan (\frac{180^{\circ} - A}{2})$ ( From equation 1 )

$= tan (90^{\circ} - \frac{A}{2})$

$= cot (\frac{A}{2})$

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t a n \frac{B + C}{2} = c o t \frac{A}{2}

t a n (\frac{C + A}{2}) = c o t \frac{B}{2}

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\frac{cos \frac{B + C}{2} cos \frac{C + A}{2} cos \frac{A + B}{2}}{sin \frac{A}{2} sin \frac{B}{2} sin \frac{C}{2}} = 1

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