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  • Solve <img alt="\cot \frac{A+B}{2} \cdot \tan \frac{A-B}{2}=?" src="https://learn.careers360.com/latex-image/?%5Ccot%20%5Cfrac%7BA&plus;B%7D%7B2%7D%20%5Ccdot%20%5Ctan%20%5Cfrac%7BA-B%7D%7B

Solve \cot \frac{A+B}{2} \cdot \tan \frac{A-B}{2}=?

Option: 1

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


Option: 2

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


Option: 3

\frac{a}{a+b}


Option: 4

None of these


Answers (1)

best_answer

 

 

Tangent Rule or Napier’s Analogy -

Tangent Rule or Napier’s Analogy

For any ΔABC , 

\\\mathrm{\tan \left(\frac{A-B}{2}\right)=\frac{a-b}{a+b} \cot \frac{C}{2}}\\\\\mathrm{\tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c} \cot \frac{A}{2}}\\\\\mathrm{\tan \left(\frac{C-A}{2}\right)=\frac{c-a}{c+a} \cot \frac{B}{2}}

-

\\\text {We know }\tan \frac{A-B}{2}=\frac{a-b}{a+b} \cot \frac{C}{2}\ .. . .(i)\\ A+B+C=\pi\\ \frac{1}{2}(A+B)=\frac{1}{2}(\pi-C)\\ \tan \frac{(A+B)}{2}=\tan \frac{(\pi-C)}{2}\\ \tan \frac{(A+B)}{2}=\cot \frac{(C)}{2}\\ \text{by equation (i)}\\ \tan \frac{A-B}{2}=\frac{a-b}{a+b} \tan \frac{(A+B)}{2}\ .. . .(i)\\ \cot \frac{A+B}{2} \cdot \tan \frac{A-B}{2}=\frac{a-b}{a+b}\\

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