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Answer 1 · 2017-11-06T17:52:54

Please refer to a Proof in the Explanation.

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

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

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

#={sinxcancel((1+cosx))}/{cosxcancel((1+cosx))},#

#=sinx/cosx,#

#=tanx.#

Hence, the Proof.

Answer 2 · 2017-11-06T17:53:01

See the demonstration below:

Start with the equation: #(tan(x)+sin(x))/(1+cos(x))=tan(x)#

Move #1+cos(x)# on the other side of the equal sign: #tan(x)+sin(x)=(1+cos(x)) cdot tan(x)#

Expand #(1+cos(x)) cdot tan(x)=tan(x)+tan(x) cdot cos(x)#

Use the equality #tan(x)=sin(x)/cos(x)#: #tan(x)+sin(x)=tan(x)+sin(x)/cos(x) cdot cos(x)#

Simplify: #tan(x)+sin(x)=tan(x)+sin(x)/cancel(cos(x)) cdot cancel(cos(x))#

You are left with: #tan(x)+sin(x)=tan(x)+sin(x)# which is true, proving that the starting equation is also true.