Prove that
(pSinB-rcosB)2 +(rsinB+pcosB)2=p2+r2

Expand the binomials, then simplify, then invoke the Pythagorean identity:

(p sin(B) - r cos(B))² + (r sin(B) + p cos(B))²

= (p² sin²(B) - 2pr sin(B) cos(B) + r ² cos²(B)) + (r ² sin²(B) + 2pr sin(B) cos(B) + p² cos²(B))

= p² (sin²(B) + cos²(B)) + r ² (cos²(B) + sin²(B))

= p² + r ²

idk

step-by-step explanation:

i think the answer is true

step-by-step explanation: