How do you prove #tanx + cotx = secx cscx#?

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Answer 1 · 2015-11-28T03:04:30

Please follow the step below

Given:
#tan x+ cot x= sec x *cscx#

Start on the right hand side, change it to #sinx# ; #cosx#

#sinx/cosx + cosx/sinx = sec x *csc x#

#color(red)([sinx/sinx])*(sinx/cosx)# + #color(blue) [cosx/cosx]*cosx/sinx# = #sec x*cscx#

#[sin^2x+cos^2x]/(sinx*cosx) = sec x *cscx#

#1/(sinx *cos x) = sec x *csc x#

#(1/sinx)(1/cosx) = secx*cscx#

#sec x *csc x = secx *csc x#

Prove completed!

*#sin^2x + cos^2x= 1#

*#1/sinx = csc x# ; #1/cosx = secx#