In a triangle ABC,with usual notations,cot le | vedclass

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(B) In a triangle $ABC$,we know that $A+B+C = \pi$,so $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.Thus,$\cot \left(\frac{A+B}{2}\right) = \cot \left

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

Vedclass · Answer

(B) In a triangle $ABC$,we know that $A+B+C = \pi$,so $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.Thus,$\cot \left(\frac{A+B}{2}ight) = \cot \left(\frac{\pi}{2} - \frac{C}{2}ight) = 	an \left(\frac{C}{2}ight)$.Now,the expression becomes $	an \left(\frac{C}{2}ight) \cdot 	an \left(\frac{A-B}{2}ight)$.Using Napier's Analogy,we have $	an \left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cot \left(\frac{C}{2}ight)$.Substituting this,we get $	an \left(\frac{C}{2}ight) \cdot \left[ \frac{a-b}{a+b} \cot \left(\frac{C}{2}ight) ight] = \frac{a-b}{a+b} \cdot \left[ 	an \left(\frac{C}{2}ight) \cdot \cot \left(\frac{C}{2}ight) ight] = \frac{a-b}{a+b} \cdot 1 = \frac{a-b}{a+b}$.

In a triangle $ABC$,with usual notations,$\cot \left(\frac{A+B}{2}\right) \cdot \tan \left(\frac{A-B}{2}\right) = $

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