Prove the following trigonometric identities.

In the given question, we need to prove $\frac{1}{\sec A+\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A-\tan A}$.

Here, we will first solve the L.H.S.

Now, using $\sec \theta=\frac{1}{\cos \theta}$ and $\tan \theta=\frac{\sin \theta}{\cos \theta}$, we get

$\frac{1}{\sec A+\tan A}-\frac{1}{\cos A}=\frac{1}{\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)}-\left(\frac{1}{\cos A}\right)$

$=\frac{1}{\left(\frac{1+\sin A}{\cos A}\right)}-\left(\frac{1}{\cos A}\right)$

$=\left(\frac{\cos A}{1+\sin A}\right)-\left(\frac{1}{\cos A}\right)$

$=\frac{\cos ^{2} A-(1+\sin A)}{(1+\sin A)(\cos A)}$

On further solving, we get

$\frac{\cos ^{2} A-(1+\sin A)}{(1+\sin A)(\cos A)}=\frac{\cos ^{2} A-1-\sin A}{(1+\sin A)(\cos A)}$

$=\frac{-\sin ^{2} A-\sin A}{(1+\sin A)(\cos A)} \quad\left(\right.$ Using $\left.\sin ^{2} \theta=1-\cos ^{2} \theta\right)$

$=\frac{-\sin A(\sin A+1)}{(1+\sin A)(\cos A)}$

$=\frac{-\sin A}{\cos A}$

$=-\tan A$

Similarly we solve the R.H.S.

Now, using $\sec \theta=\frac{1}{\cos \theta}$ and $\tan \theta=\frac{\sin \theta}{\cos \theta}$, we get

$\frac{1}{\cos A}-\frac{1}{\sec A-\tan A}=\left(\frac{1}{\cos A}\right)-\frac{1}{\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right)}$

$=\left(\frac{1}{\cos A}\right)-\frac{1}{\left(\frac{1-\sin A}{\cos A}\right)}$

$=\left(\frac{1}{\cos A}\right)-\left(\frac{\cos A}{1-\sin A}\right)$

$=\frac{(1-\sin A)-\cos ^{2} A}{(\cos A)(1-\sin A)}$

On further solving, we get

$\frac{(1-\sin A)-\cos ^{2} A}{(\cos A)(1-\sin A)}=\frac{1-\sin A-\cos ^{2} A}{(\cos A)(1-\sin A)}$

$=\frac{\sin ^{2} A-\sin A}{(\cos A)(1-\sin A)} \quad\left(\right.$ Using $\left.\sin ^{2} \theta=1-\cos ^{2} \theta\right)$

$=\frac{-\sin A(1-\sin A)}{(\cos A)(1-\sin A)}$

$=\frac{-\sin A}{\cos A}$

$=-\tan A$

So, L.H.S = R.H.S

Hence proved.