Prove that cot x + sin x / (1 + cos x) = cosec x ?

3 Answers
Narad T.
Feb 25, 2018

See the proof below

Explanation:

We need

#cotx=cosx/sinx#

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

#cscx=1/sinx#

Therefore,

#LHS=cotx+sinx/(1+cosx)#

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

#=(cosx(1+cosx)+sin^2x)/(sinx(1+cosx))#

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

#=cancel(1+cosx)/(sinxcancel(1+cosx))#

#=1/sinx#

#=cscx#

#=RHS#

#QED#

Jim G.
Feb 25, 2018

#"see explanation"#

Explanation:

#"using the "color(blue)"trigonometric identities"#

#•color(white)(x)cotx=cosx/sinx" and "cscx=1/sinx#

#•color(white)(x)sin^2x+cos^2x=1#

#"consider the left side"#

#cosx/sinx+sinx/(1+cosx)#

#"common denominator of "sinx(1+cosx)#

#=(cosx(1+cosx)+sin^2x)/(sinx(1+cosx))#

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

#=(cancel((cosx+1)))/(sinxcancel((1+cosx)))#

#=1/sinx=cscx=" right side"rArr" proved"#

Ratnaker Mehta
Feb 25, 2018

See a Proof in the Explanation.

Explanation:

#cotx+sinx/(1+cosx)#,

#=cosx/sinx+sinx/(1+cosx)#,

#={cosx(1+cosx)+sinx*sinx}/{sinx(1+cosx)}#,

#=(cosx+color(blue)(cos^2x+sin^2x))/{sinx(1+cosx)}#,

#=cancel((cosx+color(blue)1))/{sinxcancel((1+cosx))}#,

#=1/sinx#,

#=cscx," as desired!"#

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