Respuesta :

Aaronscience101

f(x)=secxcscx

let secx=g(x)

let cscx=h(x)

therefore, using leibnitz rule,

f'(x)=g'(x)*h(x)+h'(x)*g(x)

    Since derivative of secx=secxtanx

    and derivative of cscx=-cscxcotx,

f'(x)=secxtanxcscx -cscxcotxsecx

     =1/cosx*sinx/cosx*1/sinx - 1/sinx*cosx/sinx*1/cosx

     =1/cos^2x-1/sin^2x

     =(sin^2x-cos^2x)/sin^2xcos^2x / Option1

The value of f'(x) is (sin^2x - cos^2x)/sin^2x cos^2x. Thus, option A is correct.

What are trigonometric identities?

Trigonometric identities are the functions that include trigonometric functions such as sine, cosine, tangents, secant, and, cot.

Given;

f(x) = secx cscx

let secx = g(x)

let cscx = h(x)

Using Leibnitz rule,

f'(x) = g'(x)h(x) + h'(x)g(x)

Since the derivative of secx = secx tanx

derivative of cscx = -cscx cotx,

f'(x) = secx tanx cscx - cscx cotx secx

f'(x) = 1/cosx sinx/cosx 1/sinx - 1/sinx cosx/sinx 1/cosx

f'(x)  =1/cos^2x - 1/sin^2x

f'(x) = (sin^2x - cos^2x)/sin^2x cos^2x

Thus, option A is correct.

Learn more about trigonometric;

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