Question

cosx/1-tanx + sinx/1-cotx = sinx + cos x

Answer 1

write tanx=sinx/cosx and cotx=cosx/sinx, then solve it

Answer 2

there is an error in this question, this question can only be solved if one of the sign is +ve, you can't have them both negative, if you want them both negative then the answer would be -(sin(x) + cos(x))

Answer 3

\[ \sin(x)/1-\cot(x) = \frac{ \sin(x) }{ \sin(x) - \cos(x) } \]

Answer 4

similarly, cos(x)/1−tan(x)=cos(x)/[cos(x)−sin(x)]

Answer 5

now look at the denominator, what happens when you multiply them both, consider the following

Answer 6

isnt it supposed to be sin^2x and cos^2x as the numerators ?

Answer 7

(m-k)(k-m) = \[-k ^{2}-m ^{2}-km +km\]

Answer 8

yes hammy you are right but it is -1

Answer 9

-sin^2(x)-cos^2(x)=-1

Answer 10

you can figure that out just by looking at the signs

Answer 11

if it was (m-k)(m+k) then you could get a +ve answer

Answer 12

cos^2(x) at the numerator, i am just new to this editor, i;; get used to it

Answer 13

alrighty then :) wellll i got as far as ( cos^2x/cosx-sinx) +(sin^2x /sinx- cosx0

Answer 14

shoot the 0 sposed to be a closed bracket

Answer 15

see i told u hehehe

Answer 16

i need to know if this editor supports latex

Answer 17

well i cant help you with that because i do not know sorry

Answer 18

yes i think there is an error in the problem or probably the student didn't copy the question the question in the right way

Answer 19

oh well thats that

Answer 20

hey but i solved it and the ans came out alright

Answer 21

u have to take the minus common from the denominator, then it will be
\[\cos ^{2}x/(cosx-sinx) -\sin ^{2}x/(cosx-sinx)\]
then just simple adding, u will get
\[(\cos ^{2}x-\sin ^{2}x)/(cosx-sinx)\]
then apply \[a ^{2}-b ^{2}= (a+b)(a-b)\]
ie. \[(cosx+sinx)(cosx-sinx)/(cosx-sinx)\]= sinx+cosx