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