Question

prove that (1+tanA-secA)/(secA+tanA-1)=(1+secA-tanA)/(secA+tanA+1)

Answer 1

This will be a bit long too look at... but please bear with the steps :) [I'm not sure if there is a more efficient way to do this?]

First notice that:
\[\tan A=\frac{\sin A}{\cos A}, \, \,\,\, \sec A=\frac{1}{\cos A} \]
so:
\[\frac{1+\tan A-\sec A}{\sec A + \tan A-1}= \frac{1+\sec A - \tan A}{\sec A + \tan A + 1} \\ \, \\ \, \\
\large \frac{1 + \frac{\sin A}{\cos A}-\frac{1}{\cos A}}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}-1}=\frac{1+\frac{1}{\cos A}-\frac{\sin A}{\cos A}}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}+1} \\
\, \\ \, \text{Next step: I will convert all the "1"s into }\frac{\cos A}{\cos A} \text{and put everything} \\ \text{on the same denominator in one step}\\ \, \\
\large \frac{\frac{\cos A+\sin A-1}{\cos A}}{\frac{1+\sin A - \cos A}{\cos A}}=\frac{\frac{\cos A + 1 - \sin A}{\cos A}}{\frac{1+\sin A + \cos A}{\cos A}} \\
\, \\ \, \\
\frac{\cos A + \sin A -1}{1 + \sin A - \cos A}=\frac{\cos A + 1 - \sin A }{1 + \sin A + \cos A}\]

Next step, I'll cross-multiply, i.e if I have \[ \frac{a}{b}=\frac{c}{d}\], ths is the same as \[ ad = bc\]
In fact, if it seems weird that we are not keeping it strictly as LHS = RHS (since we are taking the denominators of both sides to the "other side"), observe that: \[ \frac{a\color{red}{d}}{b\color{red}{d}}=\frac{c\color{blue}{b}}{d\color{blue}{b}}\] since the denominators are equal, that's why we get \(ad = bc\), so you still maintain the form LHS = RHS.

So, cross multiplying we get:
\[ (\cos A + \sin A - 1)(1 + \sin A + \cos A)=(1 + \sin A - \cos A)(\cos A + 1- \sin A)\]

Focusing on the LHS:
\[ (\cos A + \sin A - 1)(1 + \sin A + \cos A)\\=\cancel{\cos A}+ \cos A \sin A+\cos^2 A+\cancel{\sin A} + \sin^2 A + \cos A \sin A-1-\cancel{\sin A} -\cancel{\cos A} \\
=2 \cos A \sin A+ \cos^2 A + \sin ^2 A - 1\\
=2 \cos A \sin A, \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\text{since}\, \cos^2 A+\sin^2 A=1) \]

Focusing on the RHS:
\[ (1 + \sin A - \cos A)(\cos A + 1- \sin A)\\
=\cancel{\cos A} + 1 - \cancel{\sin A}+ \cos A \sin A+ \cancel{\sin A} -\sin^2 A-\cos^2 A-\cancel{\cos A }+ \cos A \sin A\\
=1 + 2\cos A \sin A -(\sin^2 A + \cos^2 A)\\
=1 + 2\cos A \sin A - 1 \\
= 2 \cos A \sin A\]

SO!! We have shown that the left hand side (LHS) = right hand side (RHS)
Q.E.D.

Answer 2

sorry there is a cut-off
The 1st one is just "cos A" at the end where it cuts off
the 2nd one is "cos A sin A" at the end