Question

cos^2x/sin^2x +cosx secx

Answer 1

first term : use \(\Large \dfrac{\cos x}{\sin x} = \cot x\)

2nd term, use
\(\Large \dfrac{1}{\cos x } = \sec x\)

Answer 2

can you walk me through the steps please?

Answer 3

I think Cosx and 1/cosx cancel, right? so is the answer just cotx^2?

Answer 4

good start!
when we cancel terms form numerator and denominator, '1' remains!
\(\dfrac{\cancel a}{\cancel a} =1 \)

Answer 5

so, you would get
\(\Large \cot^2x+1\)
do you know any pythagorean identity in which cot^2 x is present ?

Answer 6

do you know
\(\Large \csc^2x -\cot^2 x =1 \)
??

Answer 7

oh!!! Thank you so much! I understand! It is csc^2x :) right?

Answer 8

absolutely correct! :)
good

Answer 9

you seem to be new here,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

Answer 10

oh, haha, thank you very much!:)(:

Answer 11

welcome ^_^