Here's the question you clicked on:
Hannah_Ahn
prove the identity: \[\cos x + \sin x \tan x \over \sin x \sec x \] = csc x
LHS=(cosx +sinx tanx)*cosx / sinx secx cosx =(cos^2 x +sin^2 x) / sinx =1/sinx=cscx
i think this one is easier to understand :)
I 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
because tanx cosx=sinx, secx cosx=1 multiply cos x makes LHS simpler
\[\frac{\cos(x)+\sin(x) \cdot \frac{\sin(x)}{\cos(x)}}{\sin(x) \cdot \frac{1}{ \cos(x)}} \cdot \frac{\cos(x)}{\cos(x)}=\frac{\cos^2(x)+\sin^2(x)}{\sin(x) }=\frac{1}{\sin(x)}=\csc(x)\] This is exactly what Yifan did, but I think it looks better in latex ;) Good work Yifan
Yifian: woot that was so simple! thanks! haha thanks myinninaya :)
when you meet tanx secx cscx, substitute them with sinx/cosx, 1/cosx, 1/sinx, to make the problem simpler:)