Question

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

Answer 1

LHS=(cosx +sinx tanx)*cosx / sinx secx cosx
=(cos^2 x +sin^2 x) / sinx
=1/sinx=cscx

Answer 2

i think this one is easier to understand :)

Answer 3

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

Answer 4

because tanx cosx=sinx, secx cosx=1
multiply cos x makes LHS simpler

Answer 5

\[\frac{\cos(x)+\sin(x) \cdot \frac{\sin(x)}{\cos(x)}}{\sin(x) \cdot \frac{1}{ \cos(x)}} \cdot \frac{\cos(x)}{\cos(x)}=\frac{\cos^2(x)+\sin^2(x)}{\sin(x) }=\frac{1}{\sin(x)}=\csc(x)\]
This is exactly what Yifan did, but I think it looks better in latex ;)
Good work Yifan

Answer 6

Yifian: woot that was so simple! thanks!
haha thanks myinninaya :)

Answer 7

when you meet tanx secx cscx, substitute them with sinx/cosx, 1/cosx, 1/sinx,
to make the problem simpler:)

Answer 8

got it (: