Prove : (sinA-cosA)(cosA-secA)(tanA+cotA)=1?

Answer 1

(sinA - cosA)(cosA - secA)(tanA + cotA) =? 1
(sinAcosA - tanA - cos^2A + 1)(tanA + cotA) =? 1
(sinAcosA - tanA + sin^2A)(tanA + cotA) =? 1
(sinAcosA - tanA + sin^2A)tanA + cotA(sinAcosA - tanA + sin^2A) =? 1
sin^2A - tan^2A + sin^2AtanA + cos^2A - 1 + sinAcosA =? 1
- tan^2A + sin^2AtanA + sinAcosA =? 1
- tan^2A + sin^2AtanA + tanAcos^2A =? 1
- tan^2A + tanA ≠ 1

However,
(sinA - cscA)(cosA - secA)(tanA + cotA) = 1:

(sinA - cscA)(cosA(tanA + cotA) - secA(tanA + cotA) =? 1

(sinA - cscA)(cosAtanA + cosAcotA - secAtanA - secAcotA) =? 1

(sinA - cscA)(sinA + cosAcotA - secAtanA - secAcotA) =? 1

sinA(sinA + cosAcotA - secAtanA - secAcotA) - cscA(sinA + cosAcotA - secAtanA - secAcotA) =? 1

sinAsinA + sinAcosAcotA - sinAsecAtanA - sinAsecAcotA - cscAsinA + - cscAcosAcotA - - cscAsecAtanA - - cscAsecAcotA =? 1

sin^2A + cos^2A - tan^2A - 1 - 1 - cot^2A + sec^2A + csc^2A =? 1

1 + sec^2A - (tan^2A + 1) - (1 + cot^2A) + csc^2A =? 1

1 + 0 + 0 = 1

Answer 2

Squaring sinA + cosA=sqrt3 supplies sin^2A + cos^2A + 2sinAcosA = 3 [sin^2 +cos^2 = a million] So a million + 2 sinacosA = 3 2sinAcosA = 2 Divide by using 2 sinAcosA = a million yet tanA + cotA = sinA/cosA + CosA/sinA using sinA cosA as straight forward denominator this equals [sin^2A + cos^2A] / sinA cosA = a million/sinAcosA yet from above sinAcosA = a million So tanA + cotA = a million/a million = a million

Answer 3

From what I can see, it looks like this:

(sin-cos)(cos-sec)(tan+cos)
= (sin cos - tan - cos^2 + 1)(tan+cos)
= sin^2 - tan^2 - sin cos + tan + sin cos^2 - sin - cos^3 + cos
= sin^2 - tan^2 + cos^2 (sin - cos) - sin cos + tan - sin + cos
= sin^2 (1 - 1/cos^2) + cos^2 (sin - cos) - sin (cos - 1/cos) - (sin - cos)
= sin^2 (1 - 1/cos^2) - sin cos (1 - 1/cos^2) + (sin - cos)(cos^2 - 1)
= (1 - 1/cos^2)(sin^2 - sin cos) - sin^2 (sin - cos)

From here, I'm not sure where you go with this ...

Answer 4

Sory, your question got cut off.