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

karush

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,102
The Astral plane
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

Well-known member
Jan 31, 2012
2,620
Wahiawa, Hawaii
\(\displaystyle \dfrac{tan^2(t) - 1}{tan^2(t) + 1}\)

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

Well-known member
MHB Math Helper
Aug 30, 2012
1,102
The Astral plane
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