Question

Simplify the following as much as possible

A. csc x
B. sec x
C. sin x
D. cos x

Answer 1

`\textsf{C.} \quad \sin x`

Explanation:

Given rational expression:

`(\sec^2x \csc x)/(\sec^2x + \csc^2 x)`

Rewrite the numerator and denominator of the given rational expression using the following trigonometric identities:

`\boxed{\boxed{\begin{array}{c}\underline{\sf Trigonometric\;Identities}\\\\\boxed{\sec^2 x=(1)/(\cos^2x)} \qquad \boxed{\csc^2 x=(1)/(\sin^2 x)}\qquad \boxed{\csc x=(1)/(\sin x)}\\\\\end{array}}}`

Therefore:

`=((1)/(\cos^2x) \cdot (1)/(\sin x))/((1)/(\cos^2x)+(1)/(\sin^2x))`

Multiply the fractions in the numerator, and make the denominators of the fractions in the denominator the same:

`=((1)/(\sin x\cos^2x))/((\sin^2x)/(\sin^2x\cos^2x)+(\cos^2x)/(\sin^2x\cos^2x))`

`=(\left((1)/(\sin x\cos^2x)\right))/(\left((\sin^2x+\cos^2x)/(\sin^2x\cos^2x)\right))`

`\textsf{Apply the trigonometric identity}\;\;\boxed{\sin^2x + \cos^2 x = 1}:`

`=(\left((1)/(\sin x\cos^2x)\right))/(\left((1)/(\sin^2x\cos^2x)\right))`

`\textsf{Apply\:the\:fraction\:rule}\;\;\boxed{((a)/(b))/((c)/(d))=(ad)/(bc)}:`

`=(\sin^2x\cos^2x)/(\sin x\cos^2x)`

Cancel the common factor cos²x:

`=(\sin^2x)/(\sin x)`

Simplify:

`= \sin x`

Therefore:

`\large\textsf{$(\sec^2x \csc x)/(\sec^2x + \csc^2 x)=$}\;\boxed{\boxed{\sin x}}`