Question

Proving Identities

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

Short answer

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

Step-by-step solution

Step 1. Given information

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

Step 2. Concept used

For such a query, divide the expression into two halves, LHS and RHS. Separate them and make them simpler. If two outcomes are identical, it's an identity.

Step 3. Calculation

Now, simplify LHS of$\frac{\sec u-1}{\sec u+1}=\frac{\tan u-\sin u}{\tan u+\sin u}$

We can use identity as $\sec \left(u\right)=\frac{1}{\cos \left(u\right)}$,

$\begin{array}{l}=\frac{\frac{1}{\cos u}-1}{\frac{1}{\cos u}+1}\\ =\frac{\frac{1-\cos u}{\cos u}}{\frac{1+\cos u}{\cos u}}\\ =\frac{1-\cos u}{\cos u}\times \frac{\cos u}{1+\cos u}\\ =\frac{1-\cos u}{1+\cos u}\end{array}$

Multiply and divide by $\sin u$

$\begin{array}{l}=\frac{\sin u}{\sin u}\times \frac{1-\cos u}{1+\cos u}\\ =\frac{\sin u-\sin u\times \cos u}{\sin u+\sin u\times \cos u}\end{array}$

Now multiply and divide by $\frac{1}{\cos u}$

$\begin{array}{l}=\frac{\frac{1}{\cos u}}{\frac{1}{\cos u}}\times \frac{\sin u-\sin u\times \cos u}{\sin u+\sin u\times \cos u}\\ =\frac{\frac{\sin u}{\cos u}-\sin u\times \frac{\cos u}{\cos u}}{\frac{\sin u}{\cos u}+\sin u\times \frac{\cos u}{\cos u}}\\ \because \tan u=\frac{\sin u}{\cos u}\\ =\frac{\tan u-\sin u}{\tan u+\sin u}\end{array}$

RHS of the equation is$\frac{\tan u-\sin u}{\tan u+\sin u}$

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