If tanB-C-2-x A-2- thenx -

The correct option is C b c/b + ctan B C/2 = x cot A/2⇒ x = b c/b+c. ...

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Napier’s Analogy

If tanB-C/2=x...

Question

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 =

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

If $A + B + C = π$ then, $tan \frac{A}{2} \cdot tan \frac{B}{2} + tan \frac{B}{2} \cdot tan \frac{C}{2} + tan \frac{C}{2} \cdot tan \frac{A}{2}$ =

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

Δ A B C

\frac{b - c}{b + c} = \frac{tan (\frac{B - C}{2})}{tan (\frac{B + C}{2})}

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