Proof of Trigonometry Identity: a^2b^2=sin^2θcos^2θ

[chwala]

Homework Statement

if $secθ - cosθ =a$
$cosec θ-sin θ = b$ then prove that
$a^2b^2(a^2+b^2+3) =1$

Relevant Equations

trigonometry

$a^2= \frac {sin^4θ} {cos^2θ}$
$b^2=\frac {cos^4θ} {sin^2θ}$
therefore
$a^2b^2= sin^2θ. cos^2θ$...1
and
$a^2+b^2+3= \frac {sin^6θ + cos^6θ + 3 sin^2θ cos^2θ} {sin^2θ cos^θ}$...2
multiplying 1 and 2, we get,
$sin^6θ + cos^6θ + 3 sin^2θ cos^2θ$ how do i proceed from here?

i think i got it
$sin^6θ + cos^6θ =(sin^2x)^3 + (cos^2x)^3$
using,
$a^3 + b^3 = (a+b)(a^2+b^2-ab)$
= $(sin^2x + cos^2x)(sin^4x + cos^4x-sin^2x. cos^2x)$
and using,
$a^2+b^2= (a+b)^2-2ab$
$(a^2+b^2-ab)= (a+b)^2-3ab$
thus,
=$(sin^2x + cos^2x)^2-3sin^2x. cos^2x$
therefore,

$sin^6θ + cos^6θ + 3 sin^2θ cos^2θ$ = $[(sin^2x + cos^2x)^2-3sin^2x. cos^2x]+3 sin^2θ cos^2θ$
=1
Bingo!any alternative method guys?

 

[haruspex]

I did it this way, writing s and c for sin and cos.
$ac=s^2$
$bs=c^2$
$a^2+b^2=\frac{s^4}{c^2}+\frac{c^4}{s^2}=\frac{s^6+c^6}{s^2c^2}$
Factorising sum of two cubes..
$=(s^2+c^2)\frac{s^4-s^2c^2+c^4}{s^2c^2}$
$=\frac{(s^2+c^2)^2-3s^2c^2}{s^2c^2}$
So $a^2+b^2+3=\frac{1}{s^2c^2}$
And
$a^2b^2=s^2c^2$

Having gone through that, I see what you did is essentially the same.