Question
If A, B, C are the interior angles of a triangle ABC, prove that.
$\tan\Big(\frac{\text{C+A}}{2}\Big)=\cot\frac{\text{ B}}{2}$
$\tan\Big(\frac{\text{C+A}}{2}\Big)=\cot\frac{\text{ B}}{2}$
✓
Answer
$\tan\Big(\frac{\text{C+A}}{2}\Big)=\cot\frac{\text{ B}}{2}$
In $\triangle\text{ABC},$
$\text{A + B + C}=180^\circ$
$\text{A + C}=180^\circ-\text{B}\dots(1)$
$\text{L.H.S}=\tan\Big(\frac{\text{C+A}}{2}\Big)$
$=\tan\Big(\frac{180^\circ-\text{B}}{2}\Big)$ [From (1)]
$=\tan\Big(90^\circ-\frac{\text{B}}{2}\Big)$
$=\cot\Big(\frac{\text{B}}{2}\Big)$
$=\text{R.H.S}$
In $\triangle\text{ABC},$
$\text{A + B + C}=180^\circ$
$\text{A + C}=180^\circ-\text{B}\dots(1)$
$\text{L.H.S}=\tan\Big(\frac{\text{C+A}}{2}\Big)$
$=\tan\Big(\frac{180^\circ-\text{B}}{2}\Big)$ [From (1)]
$=\tan\Big(90^\circ-\frac{\text{B}}{2}\Big)$
$=\cot\Big(\frac{\text{B}}{2}\Big)$
$=\text{R.H.S}$
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