Question

How do you use the fundamentals identities to simplify $\dfrac{{{\cos }^{2}}y}{1-\sin y}$?

Answer 1

To solve this question, we should know the trigonometric identity relation between $\sin y\And \cos y$ which states that ${{\sin }^{2}}y+{{\cos }^{2}}y=1$. We should also know the expansion formula for subtraction of squares, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. We will use these properties to simplify the given expression.

Complete step by step answer:
We know the trigonometric identity which states that ${{\sin }^{2}}y+{{\cos }^{2}}y=1$. Subtracting ${{\sin }^{2}}y$ from both sides of the identity, it can be expressed as

& \Rightarrow {{\sin }^{2}}y+{{\cos }^{2}}y-{{\sin }^{2}}y=1-{{\sin }^{2}}y \\\ & \Rightarrow {{\cos }^{2}}y=1-{{\sin }^{2}}y \\\ \end{aligned}$$ We know that the 1 is square of itself, so we can write the above expression as $${{\cos }^{2}}y={{(1)}^{2}}-{{\sin }^{2}}y$$. The RHS of this identity of the form $${{a}^{2}}-{{b}^{2}}$$, is called the subtraction of squares. We know the expansion formula for $${{a}^{2}}-{{b}^{2}}$$ is $$\left( a+b \right)\left( a-b \right)$$. Using this expansion formula for the RHS of the above identity, we get $$\begin{aligned} & \Rightarrow {{\cos }^{2}}y={{(1)}^{2}}-{{\sin }^{2}}y \\\ & \Rightarrow {{\cos }^{2}}y=\left( 1-\sin y \right)\left( 1+\sin y \right) \\\ \end{aligned}$$ We are given the given expression $$\dfrac{{{\cos }^{2}}y}{1-\sin y}$$. The numerator of the expression is $${{\cos }^{2}}y$$. As we have already shown, $${{\cos }^{2}}y$$ can also be expressed as $$\left( 1-\sin y \right)\left( 1+\sin y \right)$$. Using this in the numerator of the given expression, we get $$\begin{aligned} & \Rightarrow \dfrac{{{\cos }^{2}}y}{1-\sin y} \\\ & \Rightarrow \dfrac{\left( 1-\sin y \right)\left( 1+\sin y \right)}{\left( 1-\sin y \right)} \\\ \end{aligned}$$ As the numerator and denominator of the above expression have a common factor $$\left( 1-\sin y \right)$$, we can cancel it out. $$\begin{aligned} & \Rightarrow \dfrac{\left( 1-\sin y \right)\left( 1+\sin y \right)}{\left( 1-\sin y \right)} \\\ & \Rightarrow \left( 1+\sin y \right) \\\ \end{aligned}$$ **Hence, the expression $$\dfrac{{{\cos }^{2}}y}{1-\sin y}$$ can be written as $$1+\sin y$$ in simplified form.** **Note:** It should be noted that we can apply the above simplification only when $$\sin y\ne 1$$. Because If the $$\sin y=1$$, this means that the $$\sin y-1=0$$, in this case, the common factor of the denominator and the numerator is zero. We cannot cancel out zero as a common factor. The graph of the expression will have a hole in the graph on the values for which $$\sin y-1=0$$.