Question #def7c

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Answer 1 · 2017-04-11T22:14:13

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

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

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

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

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

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

$=2/cosx=2secx$