How do you solve $(1+sinx)/cosx+cosx/(1+sinx)=4$ in the interval $0<=x<=2pi$ ?

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Answer 1 · 2016-10-04T07:37:30

In the interval $0<=x<=2pi$ ,

$x=pi/3$ or $x=(5pi)/3$

$(1+sinx)/cosx+cosx/(1+sinx)$

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

= $(1+sinx)/cosx+(cosx-sinxcosx)/(1-sin^2x)$

= $(1+sinx)/cosx+(cosx-sinxcosx)/cos^2x$

= $(cosx+sinxcosx+cosx-sinxcosx)/cos^2x$

= $(2cosx)/cos^2x=2/cosx=2secx$

Hence, $(1+sinx)/cosx+cosx/(1+sinx)=4$

$hArr2secx=4$ or $secx=2$

As $sec(+-pi/3)=2$

and in the interval $0<=x<=2pi$ , $x=pi/3$ or $x=2pi-pi/3=(5pi)/3$

How do you solve (1+sinx)/cosx+cosx/(1+sinx)=4(1+sinx)/cosx+cosx/(1+sinx)=4 in the interval 0<=x<=2pi0<=x<=2pi?