Question

In a triangle $ABC\frac{(a+c-b)(a+b-c)}{(a+b+c)(b+c-a)}$ is equal to

• A. ${\sin }^2\frac{A}{2}$
• B. ${\cos }^2\frac{A}{2}$
• C. ${\tan }^2\frac{A}{2}$
• D. None of these

Answer 1

Answer: (C)

${\tan }^2\frac{A}{2}$

We have,

$\begin{array}{c}\frac{\left(a+c-b\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(b+c-a\right)}\\ =\frac{\left(\sin A+\sin C-\sin B\right)\left(\sin A+sinB-sinC\right)}{\left(\sin A+sinB-sinC\right)\left(sinB+\sin C-sinA\right)}\\ =\frac{\left(2\sin \left(\frac{A+C}{2}\right).\cos \left(\frac{A-C}{2}\right)-\sin B\right)\left(2\sin \left(\frac{A+B}{2}\right).\cos \left(\frac{A-B}{2}\right)-sinC\right)}{\left(2\sin \left(\frac{A+B}{2}\right).\cos \left(\frac{A-B}{2}\right)-sinC\right)\left(2\sin \left(\frac{B+C}{2}\right).\cos \left(\frac{B-C}{2}\right)-sinA\right)}\\ =\frac{\left(2\cos \frac{B}{2}.\cos \left(\frac{A-C}{2}\right)-\sin B\right)\left(2\cos \frac{C}{2}.\cos \left(\frac{A-B}{2}\right)-sinC\right)}{\left(2\cos \frac{C}{2}.\cos \left(\frac{A-B}{2}\right)-sinC\right)\left(2\cos \frac{A}{2}.\cos \left(\frac{B-C}{2}\right)-sinA\right)}\\ =\frac{\cos \frac{B}{2}\left\{\cos \left(\frac{A-C}{2}\right)-\cos \left(\frac{A+C}{2}\right)\right\}.\left\{\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right\}}{\cos \frac{A}{2}\left\{\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right\}.\left\{\cos \left(\frac{B-C}{2}\right)-\cos \left(\frac{B+C}{2}\right)\right\}}\\ =\frac{\cos \frac{B}{2}}{\cos \frac{A}{2}}.\frac{\left\{2.\sin \frac{A}{2}.\sin \frac{C}{2}\right\}\left\{2.\sin \frac{A}{2}.\sin \frac{B}{2}\right\}}{\left\{2.cos\frac{A}{2}.cos\frac{B}{2}\right\}\left\{2.\sin \frac{B}{2}.\sin \frac{C}{2}\right\}}\\ =\frac{\cos \frac{B}{2}.{\sin }^2\frac{A}{2}}{{\cos }^2\frac{A}{2}.\cos \frac{B}{2}}\\ ={\tan }^2\frac{A}{2}\\ OptionCiscorrectanswer.\end{array}$