prove that sin A + cosA = 1
secA + tanA - 1 cosecA + cotA -1
TES, you've quoted the correct version of the question, but you've taken into account the very first post lol.
We have to prove that :
$\displaystyle S=\frac{\sin(a)}{\sec(a)+\tan(a)-1}+\frac{\cos(a)}{\csc(a)+\cot(a)-1}=1$
Recall that $\displaystyle \cot=\frac{\cos}{\sin}=\frac1\tan$
$\displaystyle \sec=\frac1\cos$
$\displaystyle \csc=\frac1\sin$
this will help you make the simplifications I'll skip
$\displaystyle S=\frac{\sin(a)}{\sec(a)+\tan(a)-1}=\frac{\sin(a)}{\sec(a)+\tan(a)-1}\cdot\frac{\cot(a)}{\cot(a)}=\frac{\cos(a)}{\csc (a)+1-\cot(a)}$
Hey ! But this is very similar to $\displaystyle \frac{\cos(a)}{\csc(a)+\cot(a)-1}$
Adding the two gives :
$\displaystyle \cos(a)\left(\frac{1}{\csc(a)+1-\cot(a)}+\frac{1}{\csc(a)+\cot(a)-1}\right)$
$\displaystyle =\cos(a)\left(\frac{\csc(a)+\cot(a)-1+\csc(a)+1-\cot(a)}{(\csc(a)+1-\cot(a))(\csc(a)+\cot(a)-1)}\right)$
$\displaystyle =\cos(a)\left(\frac{2\csc(a)}{(p-q)(p+q)}\right)$
where $\displaystyle p=\csc(a)$ and $\displaystyle q=1-\cot(a)$
$\displaystyle \begin{aligned}
(p-q)(p+q)
&=p^2-q^2 \\
&=\csc^2(a)-(1-\cot(a))^2 \\
&=\csc^2(a)-\cot^2(a)+2\cot(a)-1
\end{aligned}$
But it can be very easily proved (or known) that $\displaystyle \csc^2(a)-\cot^2(a)=1$
Hence $\displaystyle (p-q)(p+q)=2\cot(a)$
And finally, we have
$\displaystyle
\frac{\sin(a)}{\sec(a)+\tan(a)-1}+\frac{\cos(a)}{\csc(a)+\cot(a)-1}=\cos(a)\cdot\frac{2\csc(a)}{2\cot(a)}$
But note that $\displaystyle \cos\cdot\csc=\cos\cdot\frac{1}{\sin}=\cot$
Hence $\displaystyle \boxed{\frac{\sin(a)}{\sec(a)+\tan(a)-1}+\frac{\cos(a)}{\csc(a)+\cot(a)-1}=1}$
Note : I've been doing all this while I'm not used at all to csc, cot, sec. What I used for simplifications is their definition in terms of cos, sin, tan. So the main difficulty for understanding is that I've been messy in my message and that I skipped some details lol