Question

Show that r+r3+r1-r2=4RcosB

Answer 1

Proved that  $r + r_{3} + r_{1} - r_{2}$ = 4R cos B

Given

To prove that $r + r_{3} + r_{1} - r_{2}$ = 4R cos B

Let us take,  $r + r_{3}$  = 4R $\frac{sin C}{2 ( \frac{sin A}{2}.\frac{sin B}{2} + \frac{cos A}{2}.\frac{cos B}{2})}$

                                = $4R sin \frac{C}{2} cos\frac{A-B}{2}$     -----> ( 1 )

                    $r_{1} - r_{2}$ = 4R $sin \frac{C}{2} }$ [ $sin \frac{A}{2} sin \frac{B}{2} + cos \frac{A}{2} cos \frac{B}{2}$ ]

                                = $4R cos \frac{C}{2} sin \frac{(A-B)}{2}$   -----> ( 2 )

Add the equation ( 1 ) and ( 2 ) to get the desired result,

                        $r + r_{3} + r_{1} - r_{2}$

                  = $4R sin \frac{C}{2} cos\frac{A-B}{2}$ + $4R cos \frac{C}{2} sin \frac{(A-B)}{2}$

                  = $4R sin [ \frac{C}{2} +\frac{A}{2} + \frac{B}{2} ]$

                  = $4R sin ( 90- \frac{B}{2} - \frac{B}{2} )$

                  = $4Rsin (90- \frac{2B}{2})$   -----> ( sin (90° - B) = cos B )

                  = 4R cos B.

Hence, proved.  

To learn more...    

brainly.in/question/12360055      

brainly.in/question/7030947

               

Answer 2

Step-by-step explanation:

I was not the final ans but u can also fallow these procedures. I hope it will help uhh.

and thank you..