Math, asked by akshithakotte4567, 6 months ago

answer ..,..............​

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Answered by LaeeqAhmed
2

\color{red}\huge{\underline{\underline{\bf GIVEN\dag}}}

  • 2 \cos( \alpha  -  \beta )  =  \sqrt{3}
  •  \sin( \alpha  +  \beta )  =  \frac{ \sqrt{3} }{2}

\color{red}\huge{\underline{\underline{\bf SOLUTION\dag}}}

2 \cos( \alpha  -  \beta )  =  \sqrt{3}

 \implies  \cos( \alpha  -  \beta ) =  \frac{ \sqrt{3} }{2}

 \bf but

\blue{ \frac{ \sqrt{3} }{2}  =  \cos(30)}

 \implies  \cos( \alpha  -  \beta )  =  \cos(30)

 \therefore  \alpha  -  \beta  = 30 \: ......(1)

\bf similarly

\sin(\alpha  + \beta ) =\frac{\sqrt{3}}{2}

\bf but

\blue{\sin(60)=\frac{\sqrt{3}}{2}}

\implies \sin(\alpha  + \beta ) =\sin(60)

\therefore  \alpha  +  \beta  = 60 \: ......(2)

\bf solving\:(1)\:and\:(2)\:we\:get

\alpha=45

\beta=15

\bf Now

\cot(\alpha  +2 \beta)=\cot(45+2(15))

\cot(\alpha  +2 \beta)=\cot(45+30)

\cot(\alpha  +2 \beta)=\cot(75)

\orange{\boxed{\therefore\cot(\alpha  +2 \beta)=2-\sqrt{3}}}

HOPE THIS HELPS!!

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