Chapter 7, Problem 19SGA

To determine

To verify: that the given equation is an identity.

Answer to Problem 19SGA

See explanation

Explanation of Solution

Given information:

  $\frac{{\sin }^{4}x-{\cos }^{4}x}{{\sin }^{2}x}=1-{\cot }^{2}x$

Calculation:

Simplify the left side of the equation into right side as follows:

  $\begin{array}{c}\frac{{\sin }^{4}x-{\cos }^{4}x}{{\sin }^{2}x}=1-{\cot }^{2}x\end{array}$

  $\begin{array}{c}\frac{{\left({\sin }^{2}x\right)}^2-{\left({\cos }^{2}x\right)}^2}{{\sin }^{2}x}=1-{\cot }^{2}x\end{array}$

  $\begin{array}{c}\frac{\left[{\sin }^{2}x-{\cos }^{2}x\right]\left[{\sin }^{2}x+{\cos }^{2}x\right]}{{\sin }^{2}x}=1-{\cot }^{2}x\left[\text{Use }{a}^2-b^2=\left(a-b\right)\left(a+b\right)\right]\end{array}$

  $\begin{array}{c}\frac{\left[{\sin }^{2}x-{\cos }^{2}x\right]\left[1\right]}{{\sin }^{2}x}=1-{\cot }^{2}x\left[\because {\sin }^{2}x+{\cos }^{2}x=1\right]\end{array}$

  $\begin{array}{c}\frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x}=1-{\cot }^{2}x\end{array}$

  $\begin{array}{c}1-\frac{{\cos }^{2}x}{{\sin }^{2}x}=1-{\cot }^{2}x\left[\because \cot x=\frac{\cos x}{\sin x}\right]\end{array}$

  $\begin{array}{c}1-{\cot }^{2}x=1-{\cot }^{2}x\end{array}$

Since the left side of the equation is transformed into right hand side. So, the equation is an identity.