Question

sinx/1-cosx+sinx/1+cosx=cscx

Answer 1

sinx/1-cosx+sinx/1+cosx=cscx

Answer 2

is it \[{\sin x \over 1-\cos x}+{\sin x \over 1+\cos x}=\csc x \]?

Answer 3

yes

Answer 4

\[={\sin x(1+\cos x)+\sin x (1-\cos x) \over 1-\cos^2 x}\]
\[={ \sin x (1+\cos x+1 -\cos x) \over \sin^2 x}\]
\[={2 \over \sin x}=2\csc x\]

Answer 5

are you sure the right hand side of the equation is cscx? because it should be 2 cscx!!
assuming the question is asking to prove the identity.

Answer 6

i also got 2cscx . are you asked to prove the equation?
if so, the equation can't be proved because it's wrong

Answer 7

no your right i miss type it thank you!

Answer 8

no problem

Answer 9

tan^2x=sec^2x-sin^2x-cos^2x
ok its the same thing

Answer 10

this is an easy one..
\[=\sec^2 x-(\sin^2 x+\cos^2 x)= \sec^2 x-1 = \tan^2 x\]
so you have both sides of the equation equal

Answer 11

tan^2x = sec^2x - (sin^2x + cos^2x)
tan^2x = sec^2x - 1
tan^2x = (1/cos^2x) -1
tan^2x = (1-cos^x)/cos^2x
tan^2x = sin^2x/cos^2x

Answer 12

sinx/cosx = tanx
so -> sin^2x/cos^2x will be equal to tan^2x

Answer 13

sec^4x-tan^4x=sec^2x+tan^2x
last one in this group

Answer 14

\[\sec^4 x-\tan^4 x=(\sec^2 x+\tan^2 x)(\sec^2 x- \tan^2 x)\]
since sec^2-tan^2x=1,
\[=\sec^2 x+\tan^2x\]

Answer 15

a fire is sighted due west of lookout A. the bearing of the fire from lookout B, 12.9 miles due south of A, is N 30^o 36' W. How far is the fire from B