Question 950607
Determine the value of sin2x if secx=a/b , where tanx<0 and secx>0 
I know that sin2x= 2sincos
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reference angle x is in quadrant IV where cos>0, sin<0,tan<0
secx=a/b
cosx=b/a
{{{sinx=-sqrt(1-cos^2(x))=-sqrt(1-(b^2/a^2))=-sqrt((a^2-b^2)/a^2)=-sqrt(a^2-b^2)/a}}}
{{{sin2x=2sinxcosx=-2(sqrt(a^2-b^2)/a)(b/a)=-sqrt(a^2-b^2)(2b/a^2)}}}