Question
Prove that $\frac{1}{(sec\theta-\tan\theta)}-\frac{1}{\cos\theta}=\frac{1}{\cos\theta}-\frac{1}{(\sec\theta+\tan\theta)}.$
✓
Answer
$\frac{1}{(\sec\theta-\tan\theta)}-\frac{1}{\cos\theta}=\frac{1}{\cos\theta}-\frac{1}{(\sec\theta+\tan\theta)}$$\text{LHS}=\frac{1}{\sec\theta-\tan\theta}-\frac{1}{\cos\theta}$
$=\frac{(\sec\theta+\tan\theta)}{(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)}-\sec\theta$ $($Multiplying the numerator and denominator by $\sec\theta+\tan\theta)$
$=\frac{\sec\theta+\tan\theta}{\sec^2\theta-\tan^2\theta}-\sec\theta$
$=\sec\theta+\tan\theta-\sec\theta$ $\big[\because\sec^2\theta-\tan^2\theta=1\big]$
$=\tan\theta$
$\text{RHS}=\frac{1}{\cos\theta}-\frac{1}{\sec\theta+\tan\theta}$
$=\sec\theta-\frac{(\sec\theta-\tan\theta)}{\sec^2\theta-\tan^2\theta}$ $($Multiplying the numerator and denominator by $\sec\theta-\tan\theta)$
$=\sec\theta+\tan\theta-\sec\theta$ $\big[\because\sec^2\theta-\tan^2\theta=1\big]$
$=\tan\theta$
$\therefore\ \text{LHS}=\text{RHS}$
Hence proved.
$=\frac{(\sec\theta+\tan\theta)}{(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)}-\sec\theta$ $($Multiplying the numerator and denominator by $\sec\theta+\tan\theta)$
$=\frac{\sec\theta+\tan\theta}{\sec^2\theta-\tan^2\theta}-\sec\theta$
$=\sec\theta+\tan\theta-\sec\theta$ $\big[\because\sec^2\theta-\tan^2\theta=1\big]$
$=\tan\theta$
$\text{RHS}=\frac{1}{\cos\theta}-\frac{1}{\sec\theta+\tan\theta}$
$=\sec\theta-\frac{(\sec\theta-\tan\theta)}{\sec^2\theta-\tan^2\theta}$ $($Multiplying the numerator and denominator by $\sec\theta-\tan\theta)$
$=\sec\theta+\tan\theta-\sec\theta$ $\big[\because\sec^2\theta-\tan^2\theta=1\big]$
$=\tan\theta$
$\therefore\ \text{LHS}=\text{RHS}$
Hence proved.
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