Chapter 14, Problem 10CLR

(a)

To determine

To prove: The equation $\sin x=\frac{\sec x}{\cot x+\tan x}$ .

Answer to Problem 10CLR

It is proved that $\sin x=\frac{\sec x}{\cot x+\tan x}$ .

Explanation of Solution

Given:

The equation is $\sin x=\frac{\sec x}{\cot x+\tan x}$ .

Calculation:

Consider the right hand side of the equation $\sin x=\frac{\sec x}{\cot x+\tan x}$ .

  $\frac{\sec x}{\cot x+\tan x}$

Simplify the expression as follows.

  $\begin{array}{c}\frac{\sec x}{\cot x+\tan x}=\frac{\frac{1}{\cos x}}{\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}}\\ =\frac{1}{\cos x}\cdot \frac{\sin x\cos x}{{\cos }^{2}x+{\sin }^{2}x}\\ =\frac{\sin x}{1}\\ =\sin x\end{array}$

Therefore, it is proved that $\sin x=\frac{\sec x}{\cot x+\tan x}$ .

(b)

To determine

To prove: The equation $\cot \left(\alpha +\beta \right)=\frac{\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }$ .

Answer to Problem 10CLR

It is proved that $\cot \left(\alpha +\beta \right)=\frac{\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }$ .

Explanation of Solution

Given:

The equation is $\cot \left(\alpha +\beta \right)=\frac{\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }$ .

Calculation:

Consider the left hand side of the equation $\cot \left(\alpha +\beta \right)=\frac{\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }$ .

  $\cot \left(\alpha +\beta \right)$

Simplify the expression as follows.

  $\begin{array}{c}\cot \left(\alpha +\beta \right)=\frac{1}{\tan \left(\alpha +\beta \right)}\\ =\frac{1}{\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }}\\ =\frac{1-\tan \alpha \tan \beta }{\tan \alpha +\tan \beta }\\ =\frac{1-\frac{1}{\cot \alpha }\cdot \frac{1}{\cot \beta }}{\frac{1}{\cot \alpha }+\frac{1}{\cot \beta }}\end{array}$

Simplify further.

  $\cot \left(\alpha +\beta \right)=\frac{\cot \alpha \cot \beta }{\cot \beta +\cot \alpha }$

Therefore, it is proved that $\cot \left(\alpha +\beta \right)=\frac{\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }$ .