What is the derivative of #f(x) = 1- sec x/ tan x#?

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Answer 1 · 2018-04-05T16:26:46

#f^'(x)=cscxcotx#

Here,

#f(x)=1-secx/tanx#

#"Applying "color(blue)" Quotient Rule"#

#f^'(x)=0-(tanxd/(dx)(secx)-secxd/(dx)(tanx))/(tanx)^2#

#=-(tanxsecxtanx-secxsec^2x)/tan^2x#

#=-secx[(tan^2x-sec^2x)/tan^2x]#

#=secx[(sec^2x-tan^2x)/tan^2x]#,where, #sec^2x-tan^2x=1#

#=(secx)/tan^2x#

#=(1/cosx)/(sin^2x/cos^2x)#

#=cosx/sin^2x#

#=1/sinx*cosx/sinx#

#=cscxcotx#

Answer 2 · 2018-04-05T16:31:10

#f^'(x)=cscxcotx#

Here,

#f(x)=1-secx/tanx#

#=1-(1/cosx)/(sinx/cosx)#

#=1-1/sinx#

#=1-cscx#

#f^'(x)=0-(-cscxcotx)#

#=cscxcotx#