if tanA=root2-1, prove that (secA.sinA tan squareA-cosecA) =(12-¡Ì5)/2

Answer 1

tanA = y/r = (√2 - 1)/1
y = √2 - 1
x = 1
so by constructing a triangle, we can find r

r^2 = 1^2 + (√2 - 1)^2
= 1 + 2 - 2√2 + 1
= 4 - 2√2

secA(sinA)(ta^2 A) - cscA
= (1/cosA)(sinA)(sin^2 A)/cos^2 A) - 1/sinA
= sin^3 A/cos^3 A - 1/sinA
= (sin^4 A - 1)/(sinAcos^3 A)
= (sin^2 A + 1)(sin^2 A - 1)/(sinAcos^3 A)
= (sin^2 A + 1)(-cos^2 A)/(sinA(cos^3 A)
= -(sin^2 A + 1)/(sinAcosA)

(sin^2 A +1)
=(√2 - 1)^2/(4-2√2) + 1
= (2 - 2√2 + 1)/(4-2√2) + (4-2√2)/(4 - 2√2)
= (7 - 4√2)/(4-2√2)
= (6-√2)/4 after rationalizing

sinAcosA = (y/r)(x/r) = xy/r^2
= (√2 - 1)(1)/(4-2√2)
= √2/4 after rationalizing

so
(secA.sinA tan squareA-cosecA)
= -(sin^2 A + 1)/(sinAcosA)
= (-6 + √2)/4 / (√2/4)
= (√2 - 6)/(√2)
= 1 - 3√2 , after rationalizing

I can't make out your (12-¡Ì5)/2
but it certainly is not what I got

I checked by finding the actual angle to be 22.5°
(tan 22.5) = √2 - 1 = .41421...
and evaluating

Answer 2

12-√5/2

Answer 3

12-√5/2