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

  • Thread starter Thread starter karush
  • Start date Start date
AI Thread Summary
The discussion focuses on proving the identity (tan² t - 1)/(sec² t) = (tan t - cot t)/(tan t + cot t). Participants explore various methods to simplify the expression, with one suggestion being to multiply the numerator and denominator by cot(t). This approach leads to a successful derivation of the identity, confirming that both sides are equal. The conversation emphasizes the desire for a concise solution, ideally in just a few steps. Ultimately, the identity is proven through algebraic manipulation and trigonometric identities.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
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?
 
Mathematics news on Phys.org
karush said:
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?
[math]\dfrac{tan^2(t) - 1}{tan^2(t) + 1}[/math]

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

-Dan
 
topsquark said:
[math]\dfrac{tan^2(t) - 1}{tan^2(t) + 1}[/math]

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

-Dan

oh cool!

[math]\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}[/math]
 
karush said:
oh cool!

[math]\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}[/math]
Or simply note that [math]cot(x) \cdot tan^2(x) = tan(x)[/math]. :)

-Dan
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top