Chapter 7.1, Problem 60E

To determine

To verify the identity:

  $\sec t\csc t\left(\tan t+\cot t\right)={\sec }^{2}t+{\csc }^{2}t$

Explanation of Solution

Given:

An identity is given as

  $\sec t\csc t\left(\tan t+\cot t\right)={\sec }^{2}t+{\csc }^{2}t$

Concept Used:

The trigonometric identities:

  $\begin{array}{l}\sec x\cos x=\frac{1}{\cos x}\\ \csc x=\frac{1}{\sin x}\\ \tan x=\frac{\sin x}{\cos x}\\ \cot x=\frac{\cos x}{\sin x}\end{array}$

Algebraic identity

  ${\left(a+b\right)}^2=a^2+2ab+b^2$

Calculation:

In order to verify the identity:

  $\sec t\csc t\left(\tan t+\cot t\right)={\sec }^{2}t+{\csc }^{2}t$

Take left hand of above identity and simplify it using the trigonometric identities.

As,

  $\begin{array}{c}\sec t\csc t\left(\tan t+\cot t\right)=\sec t\csc t\left(\frac{\sin t}{\cos t}+\frac{\cos t}{\sin t}\right)\\ =\sec t\csc t\times \frac{\sin t}{\cos t}+\sec t\csc t\times \frac{\cos t}{\sin t}\\ =\sec t\times \frac{1}{\cos t}+\csc t\times \frac{1}{\sin t}\\ =\sec t\times \sec t+\csc t\times \csc t\\ ={\sec }^{2}t+{\csc }^{2}t\\ =\text{Right hand}\end{array}$

Thus, identity is verified, i.e.

  $\sec t\csc t\left(\tan t+\cot t\right)={\sec }^{2}t+{\csc }^{2}t$