Question

In triangle $ABC$, $\angle A=\frac{\pi }{2}$, then $\tan \left(\frac{C}{2}\right)$ is equal to

• A. $\frac{a-c}{2b}$
• B. $\frac{a-b}{2c}$
• C. $\frac{a-c}{b}$
• D. $\frac{a-b}{c}$

Answer 1

Answer: (D)

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

It is given that it is right angled triangle, right angled at A.
Hence
$\sin C=\frac{c}{a}$
And
$\cos C=\frac{b}{a}$
Hence
$\tan C=\frac{c}{b}$
Or
$\tan \left(C\right)=\frac{2\tan \left(\frac{C}{2}\right)}{1-{\tan }^2\left(\frac{C}{2}\right)}$
Or
$\frac{c}{b}=\frac{2\tan \left(\frac{C}{2}\right)}{1-{\tan }^2\left(\frac{C}{2}\right)}$
$c-c{\tan }^2\left(\frac{C}{2}\right)=2b\tan \left(\frac{C}{2}\right)$
Or
$c{\tan }^2\left(\frac{C}{2}\right)+2b\tan \left(\frac{C}{2}\right)-c=0$
$\tan \left(\frac{C}{2}\right)=\frac{-2b\pm \sqrt{4c^2+4b^2}}{2c}$
Now
$b^2+c^2=a^2$
Hence
$\tan \left(\frac{C}{2}\right)=\frac{a-b}{c}$ considering $\left(\tan \right(\frac{A}{2})>0)$