Question

Which of the following expressions have value equal to four times the area of the triangle $ABC$?
(All symbols used have their usual meaning in a triangle)

• A. $rs+r_1(s-a)+r_2(s-b)+r_3(s-c)$
• B. $\frac{(a+b+c{)}^2}{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}$
• C. $(a^2+b^2-c^2)\tan B$
• D. $b^2\sin 2C+c^2\sin 2B$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $rs+r_1(s-a)+r_2(s-b)+r_3(s-c)$
B $\frac{(a+b+c{)}^2}{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}$
D $b^2\sin 2C+c^2\sin 2B$

Given : $rs+r_1(s-a)+r_2(s-b)+r_3(s-c)$
We know that
$r=\frac{\Delta }{s};r_1=\frac{\Delta }{s-a};r_2=\frac{\Delta }{s-b};r_3=\frac{\Delta }{s-c}$
So, $rs+r_1(s-a)+r_2(s-b)+r_3(s-c)=4\Delta$

$\frac{(a+b+c{)}^2}{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}$
We know that
$\tan \frac{A}{2}=\frac{\Delta }{s(s-a)}$
So, $\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}$
$=\frac{s\left[\right(s-a)+(s-b)+(s-c\left)\right]}{\Delta }=\frac{s^2}{\Delta }$
Now, $\frac{(a+b+c{)}^2}{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}$
$=\frac{4s^2}{\frac{s^2}{\Delta }}=4\Delta$

$(a^2+b^2-c^2)\tan B$
Using $a^2+b^2-c^2=2ab\cos C$, we get
$(a^2+b^2-c^2)\tan B=2ab\cos C\frac{\sin B}{\cos B}\ne 4\Delta$

$b^2\sin 2C+c^2\sin 2B$
Using sine rule, i.e. $\frac{b}{\sin B}=\frac{c}{\sin C}$, we get
$b^2\sin 2C+c^2\sin 2B=2b^2\sin C\cos C+2c^2\sin B\cos B=2bc\sin B\cos C+2cb\sin C\cos B=2bc\sin (B+C)=2bc\sin A(\because B+C=\pi -A)=4\Delta$