Chapter 7.4, Problem 46AYU

establish each identity.

$\frac{\csc \theta -1}{\cot \theta }=\frac{\cot \theta }{\csc \theta +1}$

To determine

To find: The trigonometric expression $\frac{\csc \theta -1}{\cot \theta }= \frac{\cot \theta }{\csc \theta + 1}$.

Answer to Problem 46AYU

Solution:

The trigonometric expression $\frac{\csc \theta -1}{\cot \theta }= \frac{\cot \theta }{\csc \theta + 1}$ is established.

Explanation of Solution

Given:

The trigonometric expression $\frac{\csc \theta -1}{\cot \theta }= \frac{\cot \theta }{\csc \theta + 1}$.

Calculation:

Left hand side,

$\frac{\csc \theta -1}{\cot \theta }$

By reciprocal identity for $\csc \theta$ and by Quotient identity for $\cot \theta$.

$\frac{\frac{1}{\sin \theta }-1}{\frac{\cos \theta }{\sin \theta }}$

$\frac{\frac{1-\sin \theta }{\sin \theta }}{\frac{\cos \theta }{\sin \theta }}$

Cancelling the common factor.

$\frac{1-\sin \theta }{\cos \theta }$     -------1

Right hand side,

$\frac{\cot \theta }{\csc \theta + 1}$

By quotient identity for $\cot \theta$ and reciprocal identity for $\csc \theta$.

$\frac{\frac{\cos \theta }{\sin \theta }}{\frac{1}{\sin \theta }+ 1}$

$\frac{\frac{\cos \theta }{\sin \theta }}{\frac{1+\sin \theta }{\sin \theta }}$

$\frac{\cos \theta }{1+ \sin \theta }$

Multiplying by $(1- \sin \theta )$,

$\frac{\cos \theta }{1+ \sin \theta }. \frac{1-\sin \theta }{1-\sin \theta }$

By Pythagorean identity,

$\cos \theta . \frac{1-\sin \theta }{1- si{n}^2\theta }$

$\cos \theta . \frac{1-\sin \theta }{co{s}^2\theta }$

Cancelling the common factor.

$\frac{1- \sin \theta }{\cos \theta }$     ------2

Comparing 1 and 2,

$\text{LHS }=\text{ RHS}$