Question

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

Answer 1

Simplify $\frac{\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{y}}{\mathbf{1}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{y}}$:

Firstly add and subtract ${\sin }^{2}y$ in numerator of the given trigonometry expression gives :

$\frac{{\cos }^{2}y}{1-\sin y}=\frac{{\cos }^{2}y+{\sin }^{2}y-{\sin }^{2}y}{1-\sin y}$

Then trigonometry identity ${\cos }^2\theta +{\sin }^2\theta =1$ gives :

$\begin{array}{rcl}\frac{{\cos }^{2}y}{1-\sin y}& =& \frac{1-{\sin }^{2}y}{1-\sin y}\\ & \Rightarrow & \frac{{\cos }^{2}y}{1-\sin y}=\frac{1^2-{\sin }^{2}y}{1-\sin y}\end{array}$

Use algebraic identity $a^2-b^2=\left(a+b\right)(a-b)$ :

$\begin{array}{rcl}\frac{{\cos }^{2}y}{1-\sin y}& =& \frac{\left(1-\sin y\right)\left(1+\sin y\right)}{1-\sin y}\\ & \Rightarrow & \frac{{\cos }^{2}y}{1-\sin y}=\left(1+\sin y\right)\frac{\left(1-\sin y\right)}{1-\sin y}\\ & \Rightarrow & \frac{{\cos }^{2}y}{1-\sin y}=\left(1+\sin y\right)\times 1\\ & \Rightarrow & \frac{{\cos }^{2}y}{1-\sin y}=\left(1+\sin y\right)\end{array}$

Therefore, the required simplified form of the given expression is $\frac{{\cos }^{2}y}{1-\sin y}=\left(1+\sin y\right)$.