Question

If in aright angled $△$ greatest side is ${}^{\prime }{a}^{\prime }$, then $\tan \frac{C}{2}$ equals

• A. $\frac{a-b}{c}$
• B. $\frac{a+b}{c}$
• C. $\frac{a\pm b}{c}$
• D. none of these

Answer 1

Answer: (A)

$\frac{a-b}{c}$

$\because$ Greatest side is ${}^{\prime }{a}^{\prime }$

$\angle A={90}^0\Rightarrow b^2+c^2=a^2$

$\therefore \sin C=\frac{c}{a}\Rightarrow \frac{2\tan \frac{C}{2}}{1+{\tan }^2\frac{C}{2}}=\frac{c}{a}$

$\Rightarrow 2at=c{t}^2+c;$ where $t=\tan \frac{C}{2}$

$\Rightarrow c{t}^2-2at+c=0\Rightarrow t=\frac{2a\pm \sqrt{4a^2-4c^2}}{2c}$

$\Rightarrow t=\frac{2a\pm \sqrt{4(a^2-c^2)}}{2c}=\frac{2a\pm \sqrt{4b^2}}{2c}=\frac{a\pm b}{c}$

$\therefore \tan \frac{C}{2}=\frac{a+b}{c}$ or $\frac{a-b}{c}$

If $\tan \frac{C}{2}=\frac{a+b}{c}>\frac{c}{c}=1[$ by $△$ inequality $]$

$\therefore \tan \frac{C}{2}>1\Rightarrow \frac{C}{2}>\frac{\pi }{4}\Rightarrow C>\frac{\pi }{2}$

which is impossible for right angle triangle

$\therefore \tan \frac{C}{2}=\frac{a-b}{c}$