Prove that

Solution:

According to the question,

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

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

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

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

Using the identity,

sin2A + cos2A = 1, we get,

sin A + (1 – cos A).

$\therefore \mathrm{LHS}=\frac{\sin \mathrm{A}+(1-\cos \mathrm{A})}{\sin \mathrm{A}-(1-\cos \mathrm{A})} \times \frac{\sin \mathrm{A}+(1-\cos \mathrm{A})}{\sin \mathrm{A}+(1-\cos \mathrm{A})}$

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

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

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

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

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

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

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

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

Hence, L.H.S = R.H.S