Chapter 7.1, Problem 59E

To determine

To verify the identity:

  $\frac{{\left(\sin t+\cos t\right)}^2}{\sin t\cos t}=2+\sec t\csc t$

Explanation of Solution

Given:

An identity is given as

  $\frac{{\left(\sin t+\cos t\right)}^2}{\sin t\cos t}=2+\sec t\csc t$

Concept Used:

The trigonometric identities:

  $\begin{array}{l}\sec x=\frac{1}{\cos x}\\ \csc x=\frac{1}{\sin x}\\ {\sin }^{2}x+{\cos }^{2}x=1\end{array}$

Algebraic identity

  ${\left(a+b\right)}^2=a^2+2ab+b^2$

Calculation:

In order to verify the identity:

  $\frac{{\left(\sin t+\cos t\right)}^2}{\sin t\cos t}=2+\sec t\csc t$

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

As,

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

Thus, identity is verified, i.e.

  $\frac{{\left(\sin t+\cos t\right)}^2}{\sin t\cos t}=2+\sec t\csc t$