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Hannah_Ahn
 3 years ago
prove the identity:
\[\cos x + \sin x \tan x \over \sin x \sec x \] = csc x
Hannah_Ahn
 3 years ago
prove the identity: \[\cos x + \sin x \tan x \over \sin x \sec x \] = csc x

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Yifan12879
 3 years ago
Best ResponseYou've already chosen the best response.4LHS=(cosx +sinx tanx)*cosx / sinx secx cosx =(cos^2 x +sin^2 x) / sinx =1/sinx=cscx

Yifan12879
 3 years ago
Best ResponseYou've already chosen the best response.4i think this one is easier to understand :)

Hannah_Ahn
 3 years ago
Best ResponseYou've already chosen the best response.0I don't understand, I feel so dumb.. why did you multiply deno and numerators by cos x and how you've got cos^2x+sin^2x over sinx

Yifan12879
 3 years ago
Best ResponseYou've already chosen the best response.4because tanx cosx=sinx, secx cosx=1 multiply cos x makes LHS simpler

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\cos(x)+\sin(x) \cdot \frac{\sin(x)}{\cos(x)}}{\sin(x) \cdot \frac{1}{ \cos(x)}} \cdot \frac{\cos(x)}{\cos(x)}=\frac{\cos^2(x)+\sin^2(x)}{\sin(x) }=\frac{1}{\sin(x)}=\csc(x)\] This is exactly what Yifan did, but I think it looks better in latex ;) Good work Yifan

Hannah_Ahn
 3 years ago
Best ResponseYou've already chosen the best response.0Yifian: woot that was so simple! thanks! haha thanks myinninaya :)

Yifan12879
 3 years ago
Best ResponseYou've already chosen the best response.4when you meet tanx secx cscx, substitute them with sinx/cosx, 1/cosx, 1/sinx, to make the problem simpler:)
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