Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively. If $r_1$ amd $r_2$ are the ex-radii of the traingle, then $(r_1+r_2)\tan \frac{C}{2}$ is

• A. $a$
• B. $b$
• C. $c$
• D. $ab$

Answer 1

The correct option is C $c$

$\text{FORMULA USED IN SOLVING QUESTION}$

$r_1=4R\sin \left(\frac{A}{2}\right)\cos \left(\frac{B}{2}\right)\cos \left(\frac{C}{2}\right)$

$r_2=4R\cos \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\cos \left(\frac{C}{2}\right)$

$r_3=4R\cos \left(\frac{A}{2}\right)\cos .\left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right)$

$\sin \left(A\right)\cos \left(B\right)+\cos \left(A\right)\sin \left(B\right)=\sin (A+B)$

$\sin (\frac{\pi }{2}-(C\left)\right)=\cos \left(C\right)$

$2R\sin \left(A\right)=a$

(HERE R IS THE CIRCUMRADIUS OF GIVEN TRIANGLE)

putting equation 1 and 2 in the question, we get

$(r_1+r_2)tan\left(\frac{C}{2}\right)=4R\sin \left(\frac{C}{2}\right)\left(\sin \right(\frac{A}{2}\left)\cos \right(\frac{B}{2})+\cos (\frac{A}{2}\left)\sin \right(\frac{B}{2}\left)\right)$

$=4R\sin \left(\frac{C}{2}\right)\sin \left(\pi -C2\right)$ ...(From equation 4 & $A+B+C=\pi$)

$=4R\cos \left(\frac{C}{2}\right)\sin \left(\frac{C}{2}\right)$ ...(From equation 5 )

$=c$ ...(From equation 6)