Question

Proving Identities

Verify the identity$\mathit{s}\mathit{e}\mathit{c}\mathit{v}\mathbf{-}\mathit{t}\mathit{a}\mathit{n}\mathit{v}\mathbf{=}\frac{\mathbf{1}}{\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{v}\mathbf{+}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{v}}$

Short answer

The expression$\mathit{s}\mathit{e}\mathit{c}\mathit{v}\mathbf{-}\mathit{t}\mathit{a}\mathit{n}\mathit{v}\mathbf{=}\frac{\mathbf{1}}{\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{v}\mathbf{+}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{v}}$ is an identity.

Step-by-step solution

Step 1. Given information

An expression$\sec v-\tan v=\frac{1}{\sec v+\tan v}$

Step 2. Concept used

To determine if the phrase is an identity, divide it into two parts, LHS and RHS. Make them both simpler after that. The phrase is considered to be an identity when the results are the same.

Step 3. Calculation

Now, simplify RHS of $\sec v-\tan v=\frac{1}{\sec v+\tan v}$

$\begin{array}{l}\frac{1}{\sec v+\tan v}=\frac{1}{\sec v+\tan v}\times \frac{\sec v-\tan v}{\sec v-\tan v}\\ =\frac{\mathrm{secv}-\tan v}{(\sec v+\tan v)(\sec v-\tan v)}\\ =\frac{\sec v-\tan v}{{\sec }^{2}v-{\tan }^{2}v}\\ \because {\sec }^2\theta -{\tan }^2\theta =1\\ =\frac{\sec v-\tan v}{1}\\ =\sec v-\tan v\end{array}$

LHS of the equation is$\sec v-\tan v$

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