Question

Proving Identities

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

Short answer

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

Step-by-step solution

Step 1. Given information.

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

Step 2. Concept used.

To prove the given identity, first divide it into two sections as LHS and RHS. After that, simplify them separately. If both outcomes are equal then the given expression is an identity.

Step 3. Calculation.

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

Write the expression in terms of $\sin u\&\cos u$,

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

RHS of the equation is$\frac{1}{{\cos }^{2}u-{\sin }^{2}u}$

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