8. (i) Prove that $$ ext{sec}^2 x - ext{cosec}^2 x = an^2 x - ext{cot}^2 x.$$ (ii) Given that $$y = ext{arccos} x, -1 eq x eq 1 ext{ and } 0 eq y eq ext{pi}.$$ (a) express $ ext{arcsin} x$ in terms of $y$ - Edexcel - A-Level Maths Pure - Question 1 - 2006 - Paper 6

To prove this identity, we start by rewriting the left-hand side (LHS):
$\text{sec}^2 x - \text{cosec}^2 x = \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x}$
Finding a common denominator gives us:
$= \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x}$
Now, the right-hand side (RHS):
$\tan^2 x - \text{cot}^2 x = \frac{\sin^2 x}{\cos^2 x} - \frac{\cos^2 x}{\sin^2 x}$
Again, we find a common denominator:
$= \frac{\sin^4 x - \cos^4 x}{\sin^2 x \cos^2 x}$
Recalling that ( \sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) ), and since ( \sin^2 x + \cos^2 x = 1 ):
$= \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x}$
Thus, both sides are equal, proving the identity.