How do you find the derivative of #(secx+cscx)/cscx#?

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Answer 1 · 2017-03-06T10:14:37

#d/(dx)(secx+cscx)/cscx=sec^2x#

Let us first simplify #(secx+cscx)/cscx#

= #(1/cosx+1/sin)/(1/sinx)#

= #(1/cosx+1/sin)xxsinx#

= #sinx/cosx+1#

= #tanx+1#

Hence, #d/(dx)(secx+cscx)/cscx#

= #d/(dx)(tanx+1)#

= #sec^2x+0#

= #sec^2x#

Answer 2 · 2017-03-06T10:18:01

Since, #(secx+cscx)/cscx=secx/cscx+cscx/cscx=(1/cosx)/(1/sinx) +1#

#=sinx/cosx+1=tanx+1,#

#d/dx{(secx+cscx)/cscx}=d/dx(tanx+1)=sec^2x.#