Chapter 7.4, Problem 65AYU

To determine

To find: The trigonometric expression $\frac{\sec \theta +\tan \theta }{\cot \theta +\cos \theta }= \tan \theta . \sec \theta$.

Answer to Problem 65AYU

Solution:

The trigonometric expression $\frac{\sec \theta +\tan \theta }{\cot \theta +\cos \theta }= \tan \theta . \sec \theta$ is established.

Explanation of Solution

Given:

The trigonometric expression $\frac{\sec \theta +\tan \theta }{\cot \theta +\cos \theta }= \tan \theta . \sec \theta$.

Calculation:

Left hand side,

$\frac{\sec \theta +\tan \theta }{\cot \theta +\cos \theta }$

By quotient identity for $\sec \theta$, $\tan \theta$, $\cot \theta$.

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

Adding up the quotients in the numerator and taking the common factor $\cos \theta$ in the denominator.

$\frac{\frac{1+ \sin \theta }{\cos \theta }}{\cos \theta \left(\frac{1}{\sin \theta }+ 1\right)}$

Adding up the quotients in the denominator.

$\frac{\frac{1+ \sin \theta }{\cos \theta }}{\cos \theta . \frac{\left(1+ \sin \theta \right)}{\sin \theta }}$

Cancelling the common factor $\left(1+ \sin \theta \right)$ and taking reciprocal.

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

$\tan \theta . \sec \theta$

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