Question

prove the following trigonometric identities. Include reason with calculation:

(secx+tanx)(1-sinx/cosx)=1

Answer 1

lol oops i read the question wrong

Answer 2

\[(\frac{1}{cosx}+\frac{sinx}{cosx})(1-\frac{sinx}{cosx})=\frac{1}{cosx}-\frac{1}{cosx}\frac{sinx}{cosx}+\frac{sinx}{cosx}-\frac{sinx}{cosx}\frac{sinx}{cosx}\]
\[\frac{1}{cosx}-\frac{sinx}{\cos^2x}+\frac{sinx}{cosx}-\frac{\sin^2x}{\cos^2x}\]
lets find a common denominator
\[\frac{cosx-sinx+sinxcosx-\sin^2x}{\cos^2x}\]
\[\frac{-(\sin^2x-sinxcosx)+cosx-sinx}{1-\sin^2x}=\frac{-sinx(sinx-cosx)+cosx-sinx}{(1-sinx)(1+sinx)}\]
\[=\frac{sinx(cosx-sinx)+(cosx-sinx)}{(1-sinx)(1+sinx)}=\frac{(cosx-sinx)(sinx+1)}{(1-sinx)(1+sinx)}\]
\[\frac{cosx-sinx}{1-sinx}\]

Answer 3

i'm not sure if this does =1

Answer 4

nope it does not

Answer 5

i'm going to sleep now
maybe you meant to say
(secx+tanx)((1-sinx)/cosx)=1