Question

solve the problem..........​

Answer 1

Answer:

ANSWER = 0

Step-by-step explanation:

$\frac{(sinA - sinB)(sinA+sinB) + (cosA - cosB)(cosA+cosB)}{(sinA + sinB)(cosA+cosB)}$

$\frac{sinA^{2} + cosA^{2} - (sinB^{2} + cosB^{2}) }{(sinA+sinB)(cosA + cosB)}$

now we know that sinA^2 + cosA^2 = 1 (trigonometric identity)

$\frac{1 -1 }{(sinA+sinB)(cosA + cosB)}$

ANSWER = 0

Hence proved

CHEERS MATE, PLEASE DO MARK  BRAINLIEST

Answer 2

[sin^2A-sin^2B+cos^2A-cos^2B]÷[cosA+cosB(sinA+sinB)]

=[1-(sin^2B+cos^2B)]÷[cosA+cosB(sinA+sinB)]

=[1-1]÷[cosA+cosB(sinA+sinB)]

=[0]÷[cosA+cosB(sinA+sinB)]

=0