Question

Evaluate: $\frac{1}{\sec A-\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{(\sec A+\tan A)}$.

Answer 1

Taking LHS=

$\frac{1}{\sec A-\tan A}-\frac{1}{\cos A}=\frac{1}{\sec A-\tan A}-\sec A$

$=\frac{1-(\sec A{)}^2+\sec A\times \tan A}{\sec A-\tan A}$

Multiply Numerator and Denominator by $(\sec A+\tan A)$

$=\frac{(1-(\sec A{)}^2+\sec A\times \tan A)\times (\sec A+\tan A)}{(\sec A-\tan A)\times (\sec A+\tan A)}$

We know $(\sec A{)}^2-(\tan A{)}^2=1$

Thus we have $\frac{(-(\tan A{)}^2+\sec A\times \tan A\left)\right(\sec A+\tan A)}{(\sec A-\tan A)(\sec A+\tan A)}$

$=\frac{\left(\tan A\right)(\sec A+\tan A)}{(\sec A+\tan A)}$

Again using the above identity,

$=\frac{(\sec A{)}^2-1+\sec A\times \tan A}{\sec A+\tan A}$

$=\sec A-\frac{1}{\sec A+\tan A}=$ RHS