Question

Q. If any triangle, angle C=90° prove that.
$1. sin(A-B) = \div{a²-b²}{a²+b²}$
[tex]2. cos(A-B) = \div{2ab}{c²}​

Answer 1

Answer:

proved

Step-by-step explanation:

sinA/a = sinB/b = 1/c

sinA = a/c

sinB = b/c

sin(A-B) = sinA cosB - cosA sin B = a/c( a/c ) - b/c( b/c ) = (a² - b²) /c²

sin(A-B) = (a² - b²) /( a² +b² )