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