Question #5c457

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Answer 1 · 2017-04-09T10:01:57

$LHS=(sinx + cosx) (secx+ cosecx)$

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

$=(sinx + cosx) ((sinx+cosx)/(cosxsinx))$

$= ((sinx+cosx)^2/(cosxsinx))$

$= ((sin^2x+cos^2x+2sinxcosx)/(cosxsinx))$
$= (1+2sinxcosx)/(cosxsinx)$

$= (1/(cosxsinx)+(2sinxcosx)/(cosxsinx))$

$=secxcosecx+2$

$= 2 + secx cosecx=RHS$

So this is not an equation but an identity.