Question

In a ΔABC, cot ${\displaystyle \left(\frac{\text{A - B}}{2}\right)\cdot \tan \left(\frac{\text{A + B}}{2}\right)}$ is equal to

विकल्प

• ${\displaystyle \frac{\text{a - b}}{\text{a + b}}}$

• ${\displaystyle \frac{\text{a + b}}{\text{a - b}}}$

• ${\displaystyle \frac{\text{a}\left(\text{a - b}\right)}{\text{b}\left(\text{a + b}\right)}}$

• ${\displaystyle \frac{\text{b - a}}{\text{b + a}}}$

Answer 1

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

Explanation:

cot ${\displaystyle \left(\frac{\text{A - B}}{2}\right)\cdot \tan \left(\frac{\text{A + B}}{2}\right)}$

${\displaystyle =\frac{\cot \left(\frac{\text{A - B}}{2}\right)}{\sin \left(\frac{\text{A - B}}{2}\right)}\cdot \frac{\sin \left(\frac{\text{A + B}}{2}\right)}{\sin \left(\frac{\text{A + B}}{2}\right)}}$

${\displaystyle =\frac{2\sin \left(\frac{\text{A + B}}{2}\right)\cdot \cos \left(\frac{\text{A - B}}{2}\right)}{2\cos \left(\frac{\text{A + B}}{2}\right)\cdot \cos \left(\frac{\text{A - B}}{2}\right)}}$

${\displaystyle =\frac{\sin \text{A}+\sin \text{B}}{\sin \text{A}-\sin \text{B}}}$

${\displaystyle \frac{\text{a + b}}{\text{a - b}}}$    ....[By sine rule]