Chapter 9, Problem 1CA

To determine

To find out which expressions are equivalent to $1$ .

Answer to Problem 1CA

.The following expressions are equivalent to $1$

  $\frac{{\cos }^2(-x){\tan }^{2}x}{{\sin }^2(-x)}$

  $\cos \left(\frac{\pi }{2}-x\right)\csc x$

Explanation of Solution

Given:

The following are the expressions which are given-

  $\tan x\sec x\cos x$

  ${\sin }^{2}x{\cos }^{2}x$

  $\frac{{\cos }^2(-x){\tan }^{2}x}{{\sin }^2(-x)}$

And

  $\cos \left(\frac{\pi }{2}-x\right)\csc x$

Calculation:

Let us first try out to simplify the expression-

  $\frac{{\cos }^2(-x){\tan }^{2}x}{{\sin }^2(-x)}$

Now,

  $\begin{array}{l}\frac{{\cos }^2(-x){\tan }^{2}x}{{\sin }^2(-x)}\\ ={\cot }^2(-x){\tan }^{2}x\\ ={\cot }^2(x){\tan }^{2}x\{\therefore {\cot }^2(x)\text{is an even function}}\\ =1\end{array}$

Now, evaluate the following function-

  $\begin{array}{l}\cos \left(\frac{\pi }{2}-x\right)\csc x\\ =\sin x\csc x\\ =1\end{array}$

Other trigonometric functions are not equivalent to $1$ .