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If \cot\frac{B+C}{2}.\tan \frac{B-C}{2}=x, then x equals

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

None of these


Answers (1)

best_answer

We know, \tan \frac{B-C}{2}=\frac{b-c}{b+c} \cot \frac{A}{2},

Multiply both sides by tan(A/2)

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

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

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

Hence, x=\frac{b-c}{b+c}

Posted by

Divya Prakash Singh

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