Question #db45b

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Answer 1 · 2017-01-26T13:22:10

$LHS=(1+sinx)/cosx$

Multiplying both numerator and denominator by $1-sinx$ we get

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

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

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

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

Verified

Answer 2 · 2017-01-26T13:31:18

See the Explanation

$sin^2x+cos^2x=1 rArr cos^2x=1-sin^2x$

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

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