Question

1/cosx - cosx = sin x tan x prove the identity

Answer 1

HI LANA!!!

anyway...change tanx to \(\large \frac{\sin x}{\cos x}\) that should help

Answer 2

and know that \[\sin^2 x = 1 - \cos^2 x\]

Answer 3

\[\frac{1}{\cos x}-\cos x\]
Take cos x common from both the terms
\[\cos x(\frac{1}{\cos^2 x}-1)\]
\[\cos x(\sec ^2 x-1)\]
We know
\[\sec^2 x-1=\tan^2 x\]
so
\[\cos x\times \tan^2 x\]
Can you take it from here?

Answer 4

\[\frac{1}{cosx}-\frac{\cos^2x}{cosx}=\frac{\sin^2x}{cosx}=\frac{sinx}{cosx}\times sinx\]

Answer 5

hello @lalaly :p

btw @alexeis_nicole the solution i said was for right hand side

Answer 6

@ lgbasallote yep. I'm doing that right now. but i got 1/cosx ... i'm trying to see where i went wrong for it

Answer 7

\[\sin x \tan x = \sin x (\frac{\sin x}{\cos x}) = \frac{\sin^2 x}{\cos x}\]
\[\frac{1 - \cos^2 x}{\cos x} = \frac{1}{\cos x} - \frac{\cos^2 x}{\cos x}\]

Answer 8

oh ! @lgbasallote okay, sorry i was kinda trying to figure out how you did that. now i get it. thank you !