If tan B - C-2 - x cot A-2- then x -

Tan B C/2 = x cot A/2⇒ x = b c/b+c. ...

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Napier’s Analogy

If tan B - C/...

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If $t a n \frac{B - C}{2} = x c o t \frac{A}{2}, t h e n x =$

\frac{c - a}{c + a}

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\frac{a - b}{a + b}

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\frac{b - c}{b + c}

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None of these

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t a n \frac{B - C}{2} = x c o t \frac{A}{2}, t h e n x =

△ A B C

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

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

I n △ A B C, i f x = tan (\frac{B - C}{2}) tan \frac{A}{2}, y = tan (\frac{C - A}{2}) tan \frac{B}{2}, z tan (\frac{A - B}{2}) tan \frac{C}{2}, t h e n x + y + z

Prove that : $\frac{t a n A - t a n B}{c o t B - c o t A} = \frac{t a n B}{c o t A}$

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