Question

If cosecA+cotA=b prove that (b^2+1)cosA=b^2-1

Answer 1

Given ,

b = cosecA+cotA

Making square of both sides,

b² = (1+cos²A+2cosA)/sinA

Adding 1 both sides,

b² + 1 = 2(1+cosA)/sin²A

Multiplying with cosA both sides,

(b² + 1)cosA = (2cosA+1+cos²A-sin²A)/sin²A

(b²+1)cosA  = (2cosA+1+cos²A)/sin²A - 1

(b²+1)cosA  = (1+cosA)²/sin²A - 1

(b²+1)cosA  = [(1+cosA)/sinA]² - 1

(b²+1)cosA  = (1/sinA + cosA/sinA)² - 1

(b²+1)cosA  = (cosecA + cotA)² - 1

(b²+1)cosA  = b² - 1