How do you prove that #Cosx(Tanx+Cotx) = cscx#?

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Answer 1 · 2018-07-25T06:38:50

Please see below.

#cosx(tanx+cotx)#

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

= #cosx((sin^2x+cos^2x)/(sinxcosx))#

= #cosx xx 1/(sinxcosx)#

= #1/sinx#

= #cscx#

Answer 2 · 2018-07-25T06:46:10

Please see below.

We know that,

#color(red)((1)tantheta=sintheta/costheta and cot theta=costheta/sintheta#

#color(blue)((2)sin^2theta+cos^2theta=1#

Given that,

#cosx(tanx+cotx)=cscx#

#LHS=cosx(tanx+cotx)#

#color(white)(LHS)=cosx{sinx/cosx+cosx/sinx}....tocolor(red)(Apply(1)#

#color(white)(LHS)=cosx{(sin^2x+cos^2x)/(cosxsinx)}...color(blue)(Apply(2)#

#color(white)(LHS)=cosx{1/(sinxcosx)}#

#color(white)(LHS)=1/sinx#

#color(white)(LHS)=cscx#

#:.LHS=RHS#