Question #d6553

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Answer 1 · 2017-08-04T08:32:08

We know

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

#=>1-sin^2x=cos^2x#

#=>(1-sinx)(1+sinx)=cosx*cosx#

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

Answer 2 · 2017-08-04T09:26:26

#cos^2(x)=cos^2(x)#,
#=> cos^2(x)/cos^2(x)=1#.

The identity #sin^2(x)+cos^2(x)=1# gives #cos^2(x)=1-sin^2(x)#. Then,

#cos^2(x)/(1-sin^2(x))=1#,
#=> cos^2(x)/((1-sin(x))(1+sin(x))=1#,
#=> cos(x)/(1+sin(x))=(1-sin(x))/(cos(x))#

Answer 3 · 2017-08-04T11:37:31

The answer is below.

![https://useruploads.socratic.org/v2bEtT2SSmBbI2TzOPDQ_15018465453151591985664.jpg)

Answer 4 · 2017-08-04T12:25:27

See explanation

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

L.H.S:

#:.=(1-sinx)/(cosx) xx (1+sinx)/(1+sinx)#

#:.=((1-sinx)(1+sinx))/(cosx(1+sinx)#

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

Identity:

#:.color(magenta)(sin^2x+cos^2x=1#

#:.cos^2x=1-sin^2x#

#:.(cancel(cos^2x)^(color(magenta)cosx))/(cancelcosx^color(magenta)1(1+sinx)#

#:.=color(magenta)((cosx)/(1+sinx)#

L.H.S=R.H.S. Proven