Question

Can anyone help with this identity: (1+tanx)/(1+cotx) = (1-tanx)/(cotx-1). I have already proved this identity but I was only to achieve that by manipulating both sides of the equation. They both come to tanx so LS=RS. But can anyone please solve the identity by manipulating only one side of the equation?

Answer 1

\[\Large
\frac{\tan (x)+1}{\cot (x)+1}=\frac{\frac{\sin (x)}{\cos
(x)}+1}{\frac{\cos (x)}{\sin (x)}+1}=\\\large
\Large \frac{\frac{ \sin(x) +\cos(x) } {\cos(x) }}
{\frac{\sin(x) +\cos(x)} { \sin(x)}
}= \frac{\frac{ 1 } {\cos(x) }}
{\frac{1} { \sin(x)}
}== \frac{\frac{ \cos(x)-\sin(x) } {\cos(x) }}
{\frac{cos(x)-\sin(x)} { \sin(x)}
}=\frac{1-\tan (x)}{\cot (x)-1}

\]

Answer 2

thanks so much. I also found another way to get the solution. i will post it later

Answer 3

YW