trigonometrical identities

[the_raz]

Hello, I've got a math exam tomorrow and now I'm desperately trying to study the more challenging problems in our math book and I've come across this:
Show that 1/(1-sinv)+1/(1+sinv)=2/cosv, I don't want you to give me the answer, I just want to get more clarification of what I should start with and what the main "priorities" of these kinds of equations.
I also did this one: Show that 1+tanx/(sinx+cosx)= 1/cosx, I've come to this so far-> VL= 1+tanx/sinx+cosx=1+tanx/1=1+(sinx/cosx) =-> ? How should I procceed here? (If i began correctly i.e.)
I apologize if my questions seems dumb or anything like that, and that I haven't written the equations in the "math format" (I simply don't know how to do that).
Regards

 

[soroban]

Hello, the_raz!

There is a typo in the first problem . . . it should be \(\displaystyle \cos^2\!x\)

\(\displaystyle \text{Show that: }\:\dfrac{1}{1-\sin x}+ \dfrac{1}{1+\sin x}\:=\:\dfrac{2}{\cos^2\!x}\)

Get a common denominator: multiply the first fraction by \(\displaystyle \frac{1+\sin x}{1+\sin x}\), the second by \(\displaystyle \frac{1-\sin x}{1-\sin x}\)

\(\displaystyle \displaystyle \frac{1}{1-\sin x}\cdot\frac{1+\sin x}{1+\sin x} + \frac{1}{1+\sin x}\cdot\frac{1-\sin x}{1-\sin x} \;=\;\frac{1+\sin x}{1-\sin^2\!x} + \frac{1-\sin x}{1-\sin^2\!x}\)

. . \(\displaystyle \displaystyle =\;\frac{(1 + \sin x) + (1 - \sin x)}{\cos^2\!x} \;=\;\frac{2}{\cos^2\!x}\)

\(\displaystyle \text{Show that: }\:\dfrac{1+\tan x}{\sin x+\cos x}\:=\: \dfrac{1}{\cos x}\)

\(\displaystyle \displaystyle \frac{1+\tan x}{\sin x + \cos x} \;=\;\frac{1 + \frac{\sin x}{\cos x}}{\sin x + \cos x} \)

\(\displaystyle \displaystyle \text{Multiply by }\frac{\cos x}{\cos x}\!:\;\;\frac{\cos x\left(1 + \frac{\sin x}{\cos x}\right)}{\cos x(\sin x + \cos x)} \;=\;\frac{\rlap{///////////}\cos x + \sin x}{\cos x(\rlap{///////////}\sin x + \cos x)} \;=\;\frac{1}{\cos x}\)

 

[the_raz]

Thank you for your reply, soroban, it made me understand this a lot better!