Refresh if you see display errors such as \begin{aligned}, etc.======
![\sec a + \tan a = \dfrac{\cos a}{1-\sin a}](/tex.php?f=\sec a + \tan a = \dfrac{\cos a}{1-\sin a})
focusing on the left-hand side
![\text{lhs} = \sec a + \tan a](/tex.php?f=\text{lhs} = \sec a + \tan a)
one strategy is to rewrite everything in terms of sine and cosine.since
![\sec a = \frac{1}{\cos a}](/tex.php?f=\sec a = \frac{1}{\cos a})
and
![\tan a = \frac{\sin a}{\cos a}](/tex.php?f=\tan a = \frac{\sin a}{\cos a})
, we have
![\begin{aligned} \text{lhs} & = \sec a + \tan a \\ & = \frac{1}{\cos a} + \frac{\sin a}{\cos a} \\ & = \frac{1 + \sin a}{\cos a} \end{aligned}](/tex.php?f=\begin{aligned} \text{lhs} & = \sec a + \tan a \\ & = \frac{1}{\cos a} + \frac{\sin a}{\cos a} \\ & = \frac{1 + \sin a}{\cos a} \end{aligned})
they have the same denominator, which is why i am able to combine them.that looks as simple as simple gets. can leave it alone. (read the remark at the end, because there are things you can do here, too.)======focusing on the right-hand side
![\text{rhs} = \dfrac{\cos a}{1-\sin a}](/tex.php?f=\text{rhs} = \dfrac{\cos a}{1-\sin a})
in this situation, there is one strategy we can use: multiplication by the conjugate.for the right hand side, the conjugate of
![1-\sin a](/tex.php?f=1-\sin a)
is
![1+\sin a](/tex.php?f=1+\sin a)
. we can multiply the numerator and denominator of the right hand side by this conjugate.why do this? because it will produce a difference of squares in the denominator, which may permit us to use a pythagorean identity.(another rationale is that, if you look on the left side, we have 1+sina; this can lead to you conclude that we need to multiply by 1+sina on the right side to get it there as well.)
![\begin{aligned} \text{rhs} & = \dfrac{\cos a}{1-\sin a} \cdot \frac{1+\sin a}{1+\sin a} \\ & = \dfrac{\cos a(1+\sin a)}{(1-\sin a)(1+\sin a)} \end{aligned}](/tex.php?f=\begin{aligned} \text{rhs} & = \dfrac{\cos a}{1-\sin a} \cdot \frac{1+\sin a}{1+\sin a} \\ & = \dfrac{\cos a(1+\sin a)}{(1-\sin a)(1+\sin a)} \end{aligned})
the denominator is a difference of squares expansion, in the form of
![(a-b)(a+b) = a^2 - b^2](/tex.php?f=(a-b)(a+b) = a^2 - b^2)
. doing this to the denominator gets us
![\begin{aligned} \text{rhs} & = \dfrac{\cos a(1+\sin a)}{(1-\sin a)(1+\sin a)} \\ & = \dfrac{\cos a(1+\sin a)}{1-\sin^2 a} \end{aligned}](/tex.php?f=\begin{aligned} \text{rhs} & = \dfrac{\cos a(1+\sin a)}{(1-\sin a)(1+\sin a)} \\ & = \dfrac{\cos a(1+\sin a)}{1-\sin^2 a} \end{aligned})
the denominator is a variation of the pyth. identity.with
![\sin^2 a + \cos^2 a = 1](/tex.php?f=\sin^2 a + \cos^2 a = 1)
, subtracting both sides by
![\sin^2 a](/tex.php?f=\sin^2 a)
gets us
![\cos^2 a= 1 -\sin^2 a](/tex.php?f=\cos^2 a= 1 -\sin^2 a)
. hence
![\begin{aligned} \text{rhs} & = \dfrac{\cos a(1+\sin a)}{1-\sin^2 a} \\ & = \dfrac{\cos a(1+\sin a)}{\cos^2 a} \end{aligned}](/tex.php?f=\begin{aligned} \text{rhs} & = \dfrac{\cos a(1+\sin a)}{1-\sin^2 a} \\ & = \dfrac{\cos a(1+\sin a)}{\cos^2 a} \end{aligned})
the cosines (partially) cancel out and we are left with
![\begin{aligned} \text{rhs} & = \dfrac{1+\sin a}{\cos a} \\ & = \text{lhs} \end{aligned}](/tex.php?f=\begin{aligned} \text{rhs} & = \dfrac{1+\sin a}{\cos a} \\ & = \text{lhs} \end{aligned})
======you could have also did the conjugate step for the left-hand side when you finished simplifying it all. there are usually many ways to prove an identity.i hope i'm not doing your entire homework packet or something like that (though i do understand that this one involves something different than the other questions you've posted)