Prove that $ \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{sec\theta - \tan \theta }} $ using $ se{c^2}\theta = 1 + {\tan ^2}\theta $

Hint: On the left hand side of the given equation, we have sine and cosine; on the right hand side, we have tangent and secant. So divide the numerator and denominator of the left hand side of the equation by $ \cos \theta $ . We will get a result in terms of tangent and secant as tangent is the ratio of sine and cosine; secant is the inverse of cosine. Using $ se{c^2}\theta = 1 + {\tan ^2}\theta $ , prove the given equation.

Complete step-by-step answer:
We are given to prove $ \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{sec\theta - \tan \theta }} $ .
And $ se{c^2}\theta = 1 + {\tan ^2}\theta $
Sending $ {\tan ^2}\theta $ from the right hand side to left hand side, we get
 $ se{c^2}\theta - {\tan ^2}\theta = 1 $ …….. equation (1)
Considering $ sec\theta $ as ‘a’ and $ \tan \theta $ as ‘b’, the above equation becomes $ {a^2} - {b^2} = 1 $
We already know that $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $
Therefore, $ se{c^2}\theta - {\tan ^2}\theta = \left( {sec\theta + \tan \theta } \right)\left( {sec\theta - \tan \theta } \right) $ …… equation (2)
The given equation to prove is $ \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{sec\theta - \tan \theta }} $
Considering the LHS
 $ \Rightarrow \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} $
We are now dividing the numerator and denominator of the above trigonometric expression by $ \cos \theta $
 $ \Rightarrow \dfrac{{\left( {\dfrac{{\sin \theta - \cos \theta + 1}}{{\cos \theta }}} \right)}}{{\left( {\dfrac{{\sin \theta + \cos \theta - 1}}{{\cos \theta }}} \right)}} $
 $ \Rightarrow \dfrac{{\left( {\dfrac{{\sin \theta }}{{\cos \theta }} - \dfrac{{\cos \theta }}{{\cos \theta }} + \dfrac{1}{{\cos \theta }}} \right)}}{{\left( {\dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{{\cos \theta }}{{\cos \theta }} - \dfrac{1}{{\cos \theta }}} \right)}} $
We know that the ratio of sine and cosine gives tan and the inverse of cosine is secant.
This gives,
 $ \Rightarrow \dfrac{{\tan \theta - 1 + sec\theta }}{{\tan \theta + 1 - sec\theta }} $
Writing tangent and secant terms together
 $ \Rightarrow \dfrac{{\tan \theta + sec\theta - 1}}{{\tan \theta - sec\theta + 1}} $
Now we are multiplying the numerator and denominator by $ \tan \theta - sec\theta $
 $ \Rightarrow \dfrac{{\left( {\tan \theta + sec\theta - 1} \right)\left( {\tan \theta - sec\theta } \right)}}{{\left( {\tan \theta - sec\theta + 1} \right)\left( {\tan \theta - sec\theta } \right)}} $
 $ \Rightarrow \dfrac{{\left( {\tan \theta + sec\theta } \right)\left( {\tan \theta - sec\theta } \right) - \left( {\tan \theta - sec\theta } \right)}}{{\left( {\tan \theta - sec\theta } \right)\left( {\tan \theta - sec\theta } \right) + \left( {\tan \theta - sec\theta } \right)}} $
 $ \Rightarrow \dfrac{{\left( {{{\tan }^2}\theta - se{c^2}\theta } \right) - \left( {\tan \theta - sec\theta } \right)}}{{{{\left( {\tan \theta - sec\theta } \right)}^2} + \left( {\tan \theta - sec\theta } \right)}} $ ( since from equation 2)
We know from equation 1 that $ se{c^2}\theta - {\tan ^2}\theta = 1 $ , this means $ {\tan ^2}\theta - se{c^2}\theta = - 1 $
 $ \Rightarrow \dfrac{{ - 1 - \left( {\tan \theta - sec\theta } \right)}}{{{{\left( {\tan \theta - sec\theta } \right)}^2} + \left( {\tan \theta - sec\theta } \right)}} $
 $ \Rightarrow \dfrac{{ - \left[ {1 + \left( {\tan \theta - sec\theta } \right)} \right]}}{{\left( {\tan \theta - sec\theta } \right)\left( {1 + \left( {\tan \theta - sec\theta } \right)} \right)}} $
Cancelling $ 1 + \left( {\tan \theta - sec\theta } \right) $ in the numerator and denominator, we get
 $ \Rightarrow \dfrac{{ - 1}}{{\left( {\tan \theta - sec\theta } \right)}} $
On multiplying the numerator and denominator by -1, we get
 $ \Rightarrow \dfrac{1}{{\left( {sec\theta - \tan \theta } \right)}} $
Therefore, the value of $ \dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} $ is $ \dfrac{1}{{sec\theta - \tan \theta }} $ .
Hence, proved.

Note: Here we have considered the LHS of the equation and proved it that it is equal to the RHS. We can also prove it by first considering the RHS and finding its solution. $ se{c^2}\theta - {\tan ^2}\theta = 1 $ is one of the Pythagorean identities. $ {\sin ^2}\theta + {\cos ^2}\theta = 1,\cos e{c^2}\theta - co{t^2}\theta = 1 $ are the other two Pythagorean identities.