How do you solve this?

Question

anonymous · Answer

$$\frac{ 1-sinx }{ ? cosx}+\frac{ cosx }{ 1-sinx }=2secx$$

anonymous · Answer

The question ask to show each steps to get to 2secx

anonymous · Answer

ohh and the ? is not suppose to be there

solomonzelman · Answer

$$\color{blue}{  \frac{1-\sin(x)}{?\cos(x)}+\frac{\cos(x)}{1-\sin(x)} =2\sec(x) }  $$

subtract the (value of the) second fraction from both sides

$$\color{blue}{  \frac{1-\sin(x)}{?\cos(x)}+ =2\sec(x)-\frac{\cos(x)}{1-\sin(x)} }  $$

make the right side into one simplified fraction.

solomonzelman · Answer

No plus next to the equal sign, my bad.

jdoe0001 · Answer

$\bf \cfrac{1-sin(x)}{cos(x)}+\cfrac{cos(x)}{1-sin(x)}=2sec(x)\ \quad \ \quad \
\cfrac{1-sin(x)}{cos(x)}+\cfrac{cos(x)}{1-sin(x)}\implies \cfrac{[1-sin(x)]^2+cos^2(x)}{[1-sin(x)]cos(x)}\ \quad \
\cfrac{[1-2sin(x)+{\color{blue}{ sin^2(x)]+cos^2(x)}}}{[1-sin(x)]cos(x)}$

notice those 2, so... what would you get for those?

anonymous · Answer

No the question asked you the equation got 2secx as a statement and reason. Like in geometry. Do you know how? @jdoe0001

jdoe0001 · Answer

well, yes.... so.... what would you get for $\bf sin^2(x)+cos^2(x)\quad ?$   check your trig identities