Chapter 7.1, Problem 28E

a)

To determine

To verify the identity algebraically.

  $\frac{\tan y}{\csc y}=\sec y-\cos y$

Explanation of Solution

Given:

An identity is given as

  $\frac{\tan y}{\csc y}=\sec y-\cos y$

Concept Used:

The trigonometric identities:

  $\begin{array}{l}\csc x=\frac{1}{\sin x}\\ \sec x=\frac{1}{\cos x}\\ 1-{\sin }^{2}x={\cos }^{2}x\end{array}$

Calculation:

In order to verify the identity:

  $\frac{\tan y}{\csc y}=\sec y-\cos y$

Take left hand of above identity and simplify it using the trigonometric identities.

As,

  $\begin{array}{c}\frac{\tan y}{\csc y}=\frac{\frac{\sin y}{\cos y}}{\frac{1}{\sin y}}\\ =\frac{\sin y}{\cos y}\times \sin y\\ =\frac{{\sin }^{2}y}{\cos y}\\ =\frac{1-{\cos }^{2}y}{\cos y}\\ =\frac{1}{\cos y}-\frac{{\cos }^{2}y}{\cos y}\\ =\sec y-\cos y\\ =\text{Right hand}\end{array}$

Thus, identity is verified, i.e.

  $\frac{\tan y}{\csc y}=\sec y-\cos y$

b)

To determine

To verify the identity graphically.

  $\frac{\tan y}{\csc y}=\sec y-\cos y$

Explanation of Solution

Given:

An identity is given as

  $\frac{\tan y}{\csc y}=\sec y-\cos y$

Calculation:

In order to verify the identity graphically:

  $\frac{\tan y}{\csc y}=\sec y-\cos y$

Graph of $\frac{\tan y}{\csc y}$ is constructed as

Graph of $\sec y-\cos y$ is constructed as shown:

Both graphs are same and identical.

Therefore, the identity $\frac{\tan y}{\csc y}=\sec y-\cos y$ is verified graphically.