For - ABC- prove that tan A - C2 - cot B2-

In Δ ABC , we have A + B + C = 180^o ⇒ A + C = 180^o B ⇒A + C2 = 90^o B2 ⇒ tan ( A + C2 ) = tan ( 90^o B2 ) ⇒ tan ( A + C2 ) ...

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Sign of Trigonometric Ratios in Different Quadrants

For Δ ABC, ...

Question

For $Δ A B C$ , prove that $t a n (\frac{A + C}{2}) = c o t \frac{B}{2}$ .

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Solution

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(t a n A + c o s e c B)^{2} - (c o t B - s e c A)^{2} = 2 t a n A c o t B (c o s e c A + s e c B)

Δ A B C

t a n \frac{A}{2}, t a n \frac{B}{2}, t a n \frac{C}{2}

c o t \frac{B}{2}

△ A B C

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

\frac{c}{a - b} = \frac{tan \frac{A}{2} + tan \frac{B}{2}}{tan \frac{A}{2} - tan \frac{B}{2}}

$For Δ A B C, t a n (\frac{B + C}{2}) =$ _______.

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