Question

guys pls solve.. hehe 1+ cos2x / sin2x

Answer 1

is that supposed to read \[\frac{1 + \cos (2x)}{\sin (2x)}\]
or no?

Answer 2

\[\huge{\frac{1+\cos 2x}{\sin 2x}}\]
right?

Answer 3

yes

Answer 4

\[\cos 2x=2\cos^2 x -1\]
\[\sin 2x= 2 \sin x \cos x\]
does this help?

Answer 5

the answer is cotx but i dont know how did it come up with that

Answer 6

yes it is,just put values of cos 2x and sin 2x i gave in that fraction,and cancel out terms,u will get cot x...

Answer 7

ahh yey! thank you! :D

Answer 8

welcome :)

Answer 9

wait i got confused ,, hehehe.. can u xplain it again? if thats alright with you.. :))

Answer 10

sure,which part?

Answer 11

\[\frac{1+\cos 2x}{\sin 2x}=\frac{2\cos^2 x}{2\sin x \cos x}=\frac{2cosx \cos x}{2\sin x \sin x}=\frac{\cos x}{\sin x}=cotx\]

Answer 12

i think you made a typo there...

Answer 13

yup ,sorry.
\[\frac{1+\cos 2x}{\sin 2x}=\frac{2\cos^2 x}{2\sin x \cos x}=\frac{2cosx \cos x}{2\sin x \cos x}=\frac{\cos x}{\sin x}=cotx\]

Answer 14

\[2 \cos ^{2}x\]

Answer 15

how did you get that?

Answer 16

the formula gor cos2x is 2cos^2 x -1
so 1+cos2x= 2 cos^2 x

Answer 17

i thought it should b \[2\cos ^{2}x - 1\]

Answer 18

ahh ok.. iknow now tnx alot

Answer 19

thats the formula for cos 2x,but u have +1 in the numerator also

Answer 20

how about this.. wait

Answer 21

\[\frac{ \cos ^{2} \theta - \sin ^{2}\theta}{\sin \theta \cos \theta }\]

Answer 22

another representation of cos 2x is cos^2 x-sin^2 x
so numerator directly is cos 2x
for denominator,u can uswe,sin 2x =2 sin x cos x

Answer 23

ok?

Answer 24

the answer would be then 2 cot 2x

Answer 25

tnx