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Q72.

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Found in: Page 543

### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

# Proving IdentitiesVerify the identity $\frac{{\mathrm{cos}}^{2}t+{\mathrm{tan}}^{2}t-1}{{\mathrm{sin}}^{2}t}={\mathrm{tan}}^{2}t$

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

See the step by step solution

## Step 1. Given information.

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

## 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 $\frac{{\mathrm{cos}}^{2}t+{\mathrm{tan}}^{2}t-1}{{\mathrm{sin}}^{2}t}={\mathrm{tan}}^{2}t$:

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

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

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

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

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