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Expert-verified Found in: Page 544 ### 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 $\frac{\mathbf{1}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}{\mathbf{1}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}\mathbf{=}{\mathbf{\left(}\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{x}\mathbf{-}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{x}\mathbf{\right)}}^{\mathbf{2}}$

The expression $\frac{\mathbf{1}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}{\mathbf{1}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}\mathbf{=}{\mathbf{\left(}\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{x}\mathbf{-}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{x}\mathbf{\right)}}^{\mathbf{2}}$ is an identity.

See the step by step solution

## Step 1. Given information

An expression $\frac{1-\mathrm{sin}x}{1+\mathrm{sin}x}={\left(\mathrm{sec}x-\mathrm{tan}x\right)}^{2}$

## Step 2. Concept used

Split the expression into two halves, LHS and RHS, for such a query. Separate them and simplify them. It's an identity if both results are the same.

## Step 3. Calculation

Now, simplify RHS of $\frac{1-\mathrm{sin}x}{1+\mathrm{sin}x}={\left(\mathrm{sec}x-\mathrm{tan}x\right)}^{2}$ ### Want to see more solutions like these? 