If A, B, C are the interior angles of a triangle ABC, prove that :
$t a n \frac{B + C}{2} = c o t \frac{A}{2}$

Open in App

Solution

$A + B + C = 180^{o}$
$B + C = 180^{o} - A$
$\frac{B + C}{2} = 90^{o} - \frac{A}{2}$

$t a n (\frac{B + C}{2}) = t a n (90^{o} - \frac{A}{2})$

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

Suggest Corrections

Similar questions

A, B, C

A B C

tan \frac{B + C}{2} = cot \frac{A}{2}

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

cosec (\frac{C + A}{2}) = sec (\frac{B}{2})

A, B

C

A B C

\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

A, B, C

A B C

cosec (\frac{A + B}{2}) = sec \frac{C}{2}

Join BYJU'S Learning Program