Chapter 5, Problem 7CT

To determine

To verify: The given trigonometric identity.

Explanation of Solution

Given information:

The trigonometric identity $\frac{csc\alpha +\sec \alpha }{\sin \alpha +\cos \alpha }=\cot \alpha +\tan \alpha$ .

Formula used:

The Pythagorean identity, ${\sin }^{2}x+{\cos }^{2}x=1$ .

Conversion rules, $\cot x=\frac{\cos x}{\sin x}$ , $\sec x=\frac{1}{\cos x}$ , $\tan x=\frac{\sin x}{\cos x}$ and $\csc x=\frac{1}{\sin x}$ .

Calculation:

Consider the function, $\frac{csc\alpha +\sec \alpha }{\sin \alpha +\cos \alpha }=\cot \alpha +\tan \alpha$ .

The left hand side of equation is, $\frac{csc\alpha +\sec \alpha }{\sin \alpha +\cos \alpha }$ .

Rewrite the numerator in term of sine and cosine trigonometric functions

Recall that $\sec x=\frac{1}{\cos x}$ and $\csc x=\frac{1}{\sin x}$ .

Apply it in the left hand side of the equation.

  $\begin{array}{c}\frac{csc\alpha +\sec \alpha }{\sin \alpha +\cos \alpha }=\frac{\frac{1}{\sin \alpha }+\frac{1}{\cos \alpha }}{\sin \alpha +\cos \alpha }\\ =\frac{\frac{\sin \alpha +\cos \alpha }{\sin \alpha \cos \alpha }}{\sin \alpha +\cos \alpha }\\ =\frac{\left(\sin \alpha +\cos \alpha \right)}{\sin \alpha \cos \alpha \left(\sin \alpha +\cos \alpha \right)}\\ =\frac{1}{\sin \alpha \cos \alpha }\end{array}$

Recall the Pythagorean identity, ${\sin }^{2}x+{\cos }^{2}x=1$ .

Apply it,

  $\begin{array}{c}\frac{csc\alpha +\sec \alpha }{\sin \alpha +\cos \alpha }=\frac{\left(\sin \alpha +\cos \alpha \right)}{\sin \alpha \cos \alpha \left(\sin \alpha +\cos \alpha \right)}\\ =\frac{{\sin }^2\alpha +{\cos }^2\alpha }{\sin \alpha \cos \alpha }\\ =\frac{{\sin }^2\alpha }{\sin \alpha \cos \alpha }+\frac{{\cos }^2\alpha }{\sin \alpha \cos \alpha }\\ =\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha }\end{array}$

Recall that $\cot x=\frac{\cos x}{\sin x}$ and $\tan x=\frac{\sin x}{\cos x}$ . Apply it,

  $\begin{array}{c}\frac{csc\alpha +\sec \alpha }{\sin \alpha +\cos \alpha }=\frac{{\sin }^2\alpha }{\sin \alpha \cos \alpha }+\frac{{\cos }^2\alpha }{\sin \alpha \cos \alpha }\\ =\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha }\\ =\tan \alpha +\cot \alpha \\ =\cot \alpha +\tan \alpha \end{array}$

Since, left hand side and right hand side is equal, therefore, the trigonometric identity $\frac{csc\alpha +\sec \alpha }{\sin \alpha +\cos \alpha }=\cot \alpha +\tan \alpha$ is verified.