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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sinx/1cosx+sinx/1+cosx=cscx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it \[{\sin x \over 1\cos x}+{\sin x \over 1+\cos x}=\csc x \]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[={\sin x(1+\cos x)+\sin x (1\cos x) \over 1\cos^2 x}\] \[={ \sin x (1+\cos x+1 \cos x) \over \sin^2 x}\] \[={2 \over \sin x}=2\csc x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i also got 2cscx . are you asked to prove the equation? if so, the equation can't be proved because it's wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no your right i miss type it thank you!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tan^2x=sec^2xsin^2xcos^2x ok its the same thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is an easy one.. \[=\sec^2 x(\sin^2 x+\cos^2 x)= \sec^2 x1 = \tan^2 x\] so you have both sides of the equation equal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tan^2x = sec^2x  (sin^2x + cos^2x) tan^2x = sec^2x  1 tan^2x = (1/cos^2x) 1 tan^2x = (1cos^x)/cos^2x tan^2x = sin^2x/cos^2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sinx/cosx = tanx so > sin^2x/cos^2x will be equal to tan^2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sec^4xtan^4x=sec^2x+tan^2x last one in this group

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sec^4 x\tan^4 x=(\sec^2 x+\tan^2 x)(\sec^2 x \tan^2 x)\] since sec^2tan^2x=1, \[=\sec^2 x+\tan^2x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a 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
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