Question

Proving Identities

Verify the identity$\frac{{\cos }^{2}t+{\tan }^{2}t-1}{{\sin }^{2}t}={\tan }^{2}t$

Short answer

The expression$\frac{{\cos }^{2}t+{\tan }^{2}t-1}{{\sin }^{2}t}={\tan }^{2}t$ is an identity.

Step-by-step solution

Step 1. Given information.

An expression$\frac{{\cos }^{2}t+{\tan }^{2}t-1}{{\sin }^{2}t}={\tan }^{2}t$

Step 2. Concept used.

To demonstrate that the preceding statement is an identity, divide it into two portions, LHS and RHS. Separate the two and simplify them independently when you've completed separating. If the results are equal, the given statement is an identity.

Step 3. Calculation.

Now, simplify LHS of $\frac{{\cos }^{2}t+{\tan }^{2}t-1}{{\sin }^{2}t}={\tan }^{2}t$:

Substituting 1 by ${\cos }^{2}t+{\sin }^{2}t$,

$\begin{array}{l}\frac{{\cos }^{2}t+{\tan }^{2}t-1}{{\sin }^{2}t}=\frac{{\cos }^{2}t+{\tan }^{2}t-\left({\sin }^{2}t+{\cos }^{2}t\right)}{{\sin }^{2}t}\\ =\frac{{\tan }^{2}t-{\sin }^{2}t}{{\sin }^{2}t}\\ \left(\tan x=\frac{\sin x}{\cos x}\right)\\ =\frac{{\sin }^{2}t}{{\cos }^{2}t\left({\sin }^{2}t\right)}-\frac{{\sin }^{2}t}{{\sin }^{2}t}\end{array}$

$\begin{array}{l}\frac{{\cos }^{2}t+{\tan }^{2}t-1}{{\sin }^{2}t}=\frac{1}{{\cos }^{2}t}-1\\ \left(\frac{1}{\cos x}=\sec x\right)\\ ={\sec }^{2}t-1\\ ({\sec }^{2}x+{\tan }^{2}x=1)\\ ={\tan }^{2}t\end{array}$

RHS of the equation is${\tan }^{2}t$

Both RHS and LHS are equal. Hence it is an identity