Question #4d151

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Answer 1 · 2016-11-26T04:25:16

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

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

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

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

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

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

$=2/cosx=2secx=RHS$

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