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in Trigonometry by (45 points)
Secx + tanxS/Secx- tanx  = 2+√3 / 2-√3 , then value of sin²x is?

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\(\frac{secx+tanx}{secx-tanx}\) = \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\) 

⇒ \(\frac{1+sinx}{\frac{cosx}{\frac{1-sinx}{cosx}}}\) =   \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\)  (∵ secx = \(\frac{1}{cosx}\)& tanx = \(\frac{sinx}{cosx}\))

⇒ \(\frac{1+sinx}{1-sinx}\) =   \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\) 

⇒ \(\frac{(1+sinx)+(1-sinx)}{(1+sinx)-(1-sinx)}\) =  \(\frac{(2+\sqrt{3})+(2-\sqrt{3})}{(2+\sqrt{3})-(2-\sqrt{3})}\) (∵If \(\frac{a}{b}=\frac{c}{d}\) then \(\frac{a+b}{a-b}\) = \(\frac{c+d}{c-d}\))

⇒ \(\frac{2}{2sinx}=\frac{4}{2\sqrt{3}}\) 

⇒ \(\frac{1}{sinx}=\frac{2}{\sqrt{3}}\) 

⇒ sinx = \(\frac{\sqrt{3}}{2}\) 

⇒ sin2x = \(\frac{3}{4}\) 

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