Prove (tan^2 t−1)/(sec^2 t)=(tan t−cot t)/(tan t+cot t)

[karush]

show that this is an identity

\begin{align*}\displaystyle
\frac{\tan^2 t -1}{\sec^2 t}
&=\frac{\tan t-\cot t}{\tan t+\cot t}\\
\frac{\tan^2 t -1}{\tan^2 t+1}&=
\end{align*}

OK I continued but I could not derive an equal identity

I saw a solution to this on SYM but it was many steps and got very bloated

there must be some way to do this in like 3 steps?

 

[topsquark]

show that this is an identity

\begin{align*}\displaystyle
\frac{\tan^2 t -1}{\sec^2 t}
&=\frac{\tan t-\cot t}{\tan t+\cot t}\\
\frac{\tan^2 t -1}{\tan^2 t+1}&=
\end{align*}

OK I continued but I could not derive an equal identity

I saw a solution to this on SYM but it was many steps and got very bloated

there must be some way to do this in like 3 steps?

\(\displaystyle \dfrac{tan^2(t) - 1}{tan^2(t) + 1}\)

Try multiplying the numerator and denominstor by cot(x).

-Dan

 

[karush]

\(\displaystyle \dfrac{tan^2(t) - 1}{tan^2(t) + 1}\)

Try multiplying the numerator and denominstor by cot(x).

-Dan

oh cool!

\(\displaystyle \dfrac{(tan^2(t) - 1)(\cot t)}{(tan^2(t) + 1)(\cot t)}
=\frac{\frac{\sin^\cancel{2} t}{\cos^\cancel{2} t}\frac{\cancel\cos t}{\cancel\sin t}-\frac{\cos t}{\sin t}}
{\frac{\sin^\cancel{2} t}{\cos^\cancel{2} t}\frac{\cancel\cos t}{\cancel\sin t}+\frac{\cos t}{\sin t}}
=\frac{\tan t-\cot t}{\tan t+\cot t}\)

 

[topsquark]

oh cool!

\(\displaystyle \dfrac{(tan^2(t) - 1)(\cot t)}{(tan^2(t) + 1)(\cot t)}
=\frac{\frac{\sin^\cancel{2} t}{\cos^\cancel{2} t}\frac{\cancel\cos t}{\cancel\sin t}-\frac{\cos t}{\sin t}}
{\frac{\sin^\cancel{2} t}{\cos^\cancel{2} t}\frac{\cancel\cos t}{\cancel\sin t}+\frac{\cos t}{\sin t}}
=\frac{\tan t-\cot t}{\tan t+\cot t}\)

Or simply note that \(\displaystyle cot(x) \cdot tan^2(x) = tan(x)\). :)

-Dan