Question #f5744

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Answer 1 · 2017-02-16T19:47:45

With questions like these we usually exploit the difference of squares identity:

$x^2-y^2=(x+y)(x-y)$

Multiply numerator and denominator of $LHS$ of the trig expression by $(1+sinu)$

$(1+sinu)/(1-sinu)xx(1+sinu)/(1 +sinu)$

$sinu!=1$

$=(1+2sinu+sin^2u)/(1-sin^2u)$

$"but " 1-sin^2u=cos^2u$

$=(1+2sinu+sin^2u)/cos^2u$

$=1/cos^2u+2sinu/cos^2u+sin^2u/cos^2u$

$=1/cos^2u+2color(red)(sinu/cosu)xxcolor(blue)(1/cosu)+sin^2u/cos^2u$

$=sec^2u+2color(red)(tanu)color(blue)(secu)+tan^2u$

$=(secu+tanu)^2" as required"$