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Prove that : $\frac{1}{\sec x - \tan x} - \frac{1}{\cos x} = \frac{1}{\cos x} - \frac{1}{\sec x + \tan x}$
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Sol. L.H.S. $= \frac{\sec^2 x - \tan^2 x}{\sec x - \tan x} - \sec x$ (I) ($\frac{1}{2}$ Mark)
$= \sec x + \tan x - \sec x$
$= \tan x$ (II) (1 Mark)
R.H.S. $= \sec x - \frac{\sec^2 x - \tan^2 x}{\sec x + \tan x}$ (III) ($\frac{1}{2}$ Mark)
$= \sec x - (\sec x - \tan x)$
$= \tan x$ (IV) (1 Mark)
L.H.S. $=$ R.H.S.
Alternate Solution:
Reframing, $\frac{1}{\sec x - \tan x} + \frac{1}{\sec x + \tan x} = \frac{2}{\cos x}$
LHS $= \frac{(\sec x + \tan x) + (\sec x - \tan x)}{(\sec x - \tan x) (\sec x + \tan x)}$ (I) (1 Mark)
$= \frac{2 \sec x}{\sec^2 x - \tan^2 x}$ (II) (1 Mark)
$= 2 \sec x$
$= \frac{2}{\cos x} = \text{RHS}$ (III) ($\frac{1}{2}$ Mark)
(IV) ($\frac{1}{2}$ Mark)
$= \sec x + \tan x - \sec x$
$= \tan x$ (II) (1 Mark)
R.H.S. $= \sec x - \frac{\sec^2 x - \tan^2 x}{\sec x + \tan x}$ (III) ($\frac{1}{2}$ Mark)
$= \sec x - (\sec x - \tan x)$
$= \tan x$ (IV) (1 Mark)
L.H.S. $=$ R.H.S.
Alternate Solution:
Reframing, $\frac{1}{\sec x - \tan x} + \frac{1}{\sec x + \tan x} = \frac{2}{\cos x}$
LHS $= \frac{(\sec x + \tan x) + (\sec x - \tan x)}{(\sec x - \tan x) (\sec x + \tan x)}$ (I) (1 Mark)
$= \frac{2 \sec x}{\sec^2 x - \tan^2 x}$ (II) (1 Mark)
$= 2 \sec x$
$= \frac{2}{\cos x} = \text{RHS}$ (III) ($\frac{1}{2}$ Mark)
(IV) ($\frac{1}{2}$ Mark)