Tan×+cosx÷1+sinx?

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Answer 1 · 2018-05-16T22:17:49

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

I'm assuming you meant #cosx-:(1+sinx)#, as #cosx-:1# is trivial.

When we have #1+sinx# in the denominator, we want to amplify the fraction by #1-sinx# to get #1-sin^2x=cos^2x#.

#:. E(x)=tanx+cosx/(1+sinx)*color(red)((1-sinx)/(1-sinx))#

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

Simplifying the #cosx#:

#E(x)=tanx+(1-sinx)/cosx#

We know, by definition, that #tanx=sinx/cosx#.

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

#=>color(red)(E(x)=secx)#