Chapter 7, Problem 11CT

To determine

To prove: The identity $\frac{\csc \theta +\cot \theta }{\sec \theta +\tan \theta }=\frac{\sec \theta -\tan \theta }{\csc \theta -\cot \theta }$.

Explanation of Solution

Given information:

Instruction is to establish the identity $\frac{\csc \theta +\cot \theta }{\sec \theta +\tan \theta }=\frac{\sec \theta -\tan \theta }{\csc \theta -\cot \theta }$

Formula used:

1) ${\csc }^2\theta -{\cot }^2\theta =1$

2) ${\sec }^2\theta -{\tan }^2\theta =1$

Proof:

Consider L.H.S of identity,

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

Rationalize the numerator,

$\Rightarrow \frac{\csc \theta +\cot \theta }{\sec \theta +\tan \theta }\cdot \frac{\csc \theta -\cot \theta }{\csc \theta -\cot \theta }$

By simplifying it,

$\Rightarrow \frac{{\csc }^2\theta -{\cot }^2\theta }{\left(\sec \theta +\tan \theta \right)\left(\csc \theta -\cot \theta \right)}$

By using ${\csc }^2\theta -{\cot }^2\theta =1$,

$\Rightarrow \frac{1}{\left(\sec \theta +\tan \theta \right)\left(\csc \theta -\cot \theta \right)}$

Multiply the numerator and denominator by $\sec \theta -\tan \theta$,

$\Rightarrow \frac{1}{\left(\sec \theta +\tan \theta \right)\left(\csc \theta -\cot \theta \right)}\cdot \frac{\sec \theta -\tan \theta }{\sec \theta -\tan \theta }$

$\Rightarrow \frac{\sec \theta -\tan \theta }{\left({\sec }^2\theta -{\tan }^2\theta \right)\left(\csc \theta -\cot \theta \right)}$

By using ${\sec }^2\theta -{\tan }^2\theta =1$,

$\Rightarrow \frac{\sec \theta -\tan \theta }{\left(1\right)\left(\csc \theta -\cot \theta \right)}$

$\Rightarrow \frac{\sec \theta -\tan \theta }{\csc \theta -\cot \theta }$

= R.H.S.

Hence, the identity $\frac{\csc \theta +\cot \theta }{\sec \theta +\tan \theta }=\frac{\sec \theta -\tan \theta }{\csc \theta -\cot \theta }$ is proved.