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"#