Question

Proving Identities

Verify the identity$\frac{{(\sin x+\cos x)}^2}{{\sin }^{2}x-{\cos }^{2}x}=\frac{{\sin }^{2}x-{\cos }^{2}x}{{(\sin x-\cos x)}^2}$

Short answer

The expression$\frac{{(\sin x+\cos x)}^2}{{\sin }^{2}x-{\cos }^{2}x}=\frac{{\sin }^{2}x-{\cos }^{2}x}{{(\sin x-\cos x)}^2}$ is an identity.

Step-by-step solution

Step 1. Given information.

An expression$\frac{{(\sin x+\cos x)}^2}{{\sin }^{2}x-{\cos }^{2}x}=\frac{{\sin }^{2}x-{\cos }^{2}x}{{(\sin x-\cos x)}^2}$

Step 2. Concept used.

By dividing the equation into LHS and RHS, we can check it. Then, separately, simplify both words. The expression is an identity if both responses are the same.

Step 3. Calculation.

Now, simplify LHS of $\frac{{(\sin x+\cos x)}^2}{{\sin }^{2}x-{\cos }^{2}x}=\frac{{\sin }^{2}x-{\cos }^{2}x}{{(\sin x-\cos x)}^2}$:

$$\begin{gathered} =\frac{{(\sin x+\cos x)}^2}{{\sin }^{2}x-{\cos }^{2}x}=\frac{{(\sin x+\cos x)}^2}{(\sin x+\cos x)(\sin x-\cos x)} \\ =\left(\frac{\sin x+\cos x}{\sin x-\cos x}\right) \\ =\frac{(\sin x+\cos x)}{(\sin x-\cos x)}\times \frac{(\sin x-\cos x)}{(\sin x-\cos x)} \\ =\frac{{\sin }^{2}x-{\cos }^{2}x}{{(\sin x-\cos x)}^2} \end{gathered}$$

RHS of the equation is$\frac{{\sin }^{2}x-{\cos }^{2}x}{{(\sin x-\cos x)}^2}$

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