Can anyone help me with (sec(x)+tan(x)/sec(x)-tan(x))=(sec(x)+tan(x))^2

Question

aravindg · Answer

Open up (sec(x)+tan(x))^2

aravindg · Answer

Then use the identity sec^2 x+tan^2 x=1

anonymous · Answer

alright so i now have the left side = to 1

anonymous · Answer

do i multiply by secx+tanx  top and bottom on the left?

anonymous · Answer

Yes

anonymous · Answer

sec^2 x-tan^2 x=1 and not +

anonymous · Answer

i then have sec(x)^2+2sec(x)tan(x)+ tan(x)^2 / sec^2-tan^2

anonymous · Answer

It is straight forward now
$$
\frac{(\tan (x)+\sec (x)) (\tan (x)+\sec (x))}{(\sec (x)-\tan (x)) (\tan
   (x)+\sec (x))}=\frac{(\tan (x)+\sec (x))^2}{\sec ^2(x)-\tan
   ^2(x)}=\ (\tan (x)+\sec (x))^2
$$

anonymous · Answer

$$
\frac{(\tan (x)+\sec (x)) (\tan (x)+\sec (x))}{(\sec (x)-\tan (x)) (\tan
   (x)+\sec (x))}=\frac{(\tan (x)+\sec (x))^2}{\sec ^2(x)-\tan
   ^2(x)}=\ (\tan (x)+\sec (x))^2
$$

anonymous · Answer

thank you i had foiled the problem and gotten confused

anonymous · Answer

It is pretty simple