In ABC, tan A = \(\frac{BC}{AB}\) = 1
BC = AB
Let AB = BC = k, where k is a positive number

Now, AC = \(\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \)
= \(\sqrt { { (k) }^{ 2 }+{ (k) }^{ 2 } } =k\sqrt { 2 } \)
Therefore,
\(sinA=\frac { BC }{ AC } =\frac { 1 }{ \sqrt { 2 } } \) and
\(cosA=\frac { AB }{ Ac } =\frac { 1 }{ \sqrt { 2 } } \)
So, \(2sinAcosA=2\left[ \frac { 1 }{ \sqrt { 2 } } \right] \left[ \frac { 1 }{ \sqrt { 2 } } \right] =1\), which is the required value

LHS = sec A (1 - sin A) (sec A + tan A)
= \(\left[ \frac { 1 }{ cosA } \right] (1-sinA)\left[ \frac { 1 }{ cosA } +\frac { sinA }{ cosA } \right] \)
= \(\frac { (1-sinA)(1+cosA) }{ { cos }^{ 2 }A } \)
= \(\frac { 1-{ sin }^{ 2 }A }{ { cos }^{ 2 }A } \)
= \(\frac { { cos }^{ 2 }A }{ { cos }^{ 2 }A } \) = 1 = RHS

Since we will apply the identity involving sec θ and tan θ, let us first convert the LHS (of the identity we need to prove) in terms of sec θ and tan θ by dividing numerator and denominator by cos θ.
LHS = \(\frac { sin\theta -cos\theta +1 }{ sin\theta +cos\theta -1 } =\frac { tan\theta -1+sec\theta }{ tan\theta +1-sec\theta } \)
= \(\frac { (tan\theta +sec\theta )-1 }{ (tan\theta -sec\theta )+1 } \)
= \(\frac { \{ (tan\theta +sec\theta )-1\} (tan\theta -sec\theta ) }{ \{ tan\theta -sec\theta )+1\} (tan\theta -sec\theta ) } \)
= \(\frac { ({ tan }^{ 2 }\theta -{ sec }^{ 2 }\theta )-(tan\theta -sec\theta ) }{ (tan\theta -sec\theta +1)(tan\theta -sec\theta ) } \)
= \(\frac { -1-tan\theta +sec\theta }{ (tan\theta -sec\theta +1)(tan\theta -sec\theta ) } \)
= \(\frac { -1 }{ tan\theta -sec\theta } \)
= \(\frac { 1 }{ sec\theta -tan\theta } \)

Let us first draw a right ΔABC.
Now, we know that tan A = \(\frac { BC }{ AB } =\frac { 4 }{ 3 } \)
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.

Now, by using the Pythagoras theorem, we have
AC² = AB² + BC²
= (4k)² + (3k)² = 25k²
So, AC = 5k
Now, we can write all the trigonometric ratios using their definitions.
\(sinA=\frac { BC }{ AC } =\frac { 4k }{ 5k } =\frac { 4 }{ 5 } \)
\(cosA=\frac { AB }{ AC } =\frac { 3k }{ 5k } =\frac { 3 }{ 5 } \)
Therefore, \(cotA=\frac { 1 }{ tanA } =\frac { 3 }{ 4 } \)
\(cosecA=\frac { 1 }{ sinA } =\frac { 5 }{ 4 } \) and,
\(secA=\frac { 1 }{ cosA } =\frac { 5 }{ 3 } \)