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

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Solution

$t a n (\frac{C + A}{2}) = c o t (\frac{B}{2})$
Prove $\Rightarrow$ In a $Δ$
$A + B + C = 180^{\circ} A + C = 180 - B \frac{C + A}{2} = 90 - \frac{B}{2}$
$t a n (\frac{C + A}{2}) = t a n (90 - \frac{B}{2}) = c o t (\frac{B}{2})$ $[∵ t a n (90 - θ) = c o t θ]$

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A, B, C

A B C

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

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

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