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, $(secx+cscx)/cscx=secx/cscx+cscx/cscx=(1/cosx)/(1/sinx) +1$

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

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

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