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,

$cosx(tanx+cotx)=cscx$

$LHS=cosx(tanx+cotx)$

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

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

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

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

$color(white)(LHS)=cscx$

$:.LHS=RHS$

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