## anonymous 5 years ago sinx/1-cosx+sinx/1+cosx=cscx

1. anonymous

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

2. anonymous

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

3. anonymous

yes

4. anonymous

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

5. anonymous

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.

6. anonymous

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

7. anonymous

no your right i miss type it thank you!

8. anonymous

no problem

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

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