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

• MHB
• karush
In summary, the given equation is a trigonometric identity that relates tangent, cotangent, and secant functions. It can be proven using basic trigonometric identities and can be used to solve trigonometric equations. Understanding and memorizing trigonometric identities is important for solving complex equations and has applications in various fields. Practice and understanding the derivation of identities can aid in memorization.
karush
Gold Member
MHB
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?

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

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

-Dan

topsquark said:
$$\displaystyle \dfrac{tan^2(t) - 1}{tan^2(t) + 1}$$

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

-Dan

oh cool!

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

karush said:
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

## 1. What is the given equation asking to prove?

The given equation is asking to prove that (tan^2 t−1)/(sec^2 t)=(tan t−cot t)/(tan t+cot t).

## 2. What are the steps to prove this equation?

To prove this equation, we can start by using the identity tan^2 t + 1 = sec^2 t. Then, we can multiply both sides of the equation by (tan t + cot t) to get (tan^2 t − 1)(tan t + cot t) = (tan t − cot t)(sec^2 t). From there, we can simplify and use the identity tan t = sin t/cos t and cot t = cos t/sin t to get the final simplified equation.

## 3. Can this equation be proven using a trigonometric identity?

Yes, this equation can be proven using the trigonometric identity tan^2 t + 1 = sec^2 t.

## 4. Is there a specific domain for which this equation holds true?

This equation holds true for all real values of t except for t = (2n+1)π/2 where n is any integer. This is because at these values, sec^2 t is undefined.

## 5. How can this equation be applied to real-life situations?

This equation can be applied in various fields such as physics, engineering, and astronomy. It can be used to solve problems involving triangles and circular motion, among others.

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