Question

1/secA -tanA-1/cos A =1/cos A - 1/secA +tanA

Answer 1

$\dfrac{1}{\sec A -\tan A}-\dfrac{1}{\cos A}=\dfrac{1}{cos A}-\dfrac{1}{\sec A+\tan A}$, proved.

Step-by-step explanation:

Prove that, $\dfrac{1}{\sec A -\tan A}-\dfrac{1}{\cos A}=\dfrac{1}{cos A}-\dfrac{1}{\sec A+\tan A}$.

L.H.S.$=\dfrac{1}{\sec A -\tan A}-\dfrac{1}{\cos A}$

$=\dfrac{1}{\sec A -\tan A}\times \dfrac{\sec A +\tan A}{\sec A +\tan A}-\dfrac{1}{\cos A}$

$=\dfrac{\sec A +\tan A}{\sec^2 A -\tan^2 A}-\dfrac{1}{\cos A}$

$=\sec A +\tan A-\sec A$

[ ∵ $\sec^2 A -\tan^2 A=1$ and $\sec A=\dfrac{1}{\cos A}$]

=\tan A

R.H.S.$=\dfrac{1}{cos A}-\dfrac{1}{\sec A+\tan A}$

$=\dfrac{1}{cos A}-\dfrac{1}{\sec A+\tan A}\times \dfrac{\sec A-\tan A}{\sec A-\tan A}$

$=\dfrac{1}{cos A}-\dfrac{\sec A-\tan A}{\sec^2 A-\tan^2 A}$

$=\sec A -(\sec A-\tan A)$

$=\sec A -\sec A+\tan A$

$=\tan A$

∵ L.H.S. = R.H.S. $=\tan A$, proved,