Math, asked by kunalmehra711, 1 year ago

cot (-600°) find the value​

Answers

Answered by AbhijithPrakash
16

Answer:

\displaystyle\cot \left(-600^{\circ \:}\right)=-\frac{\sqrt{3}}{3}\quad \begin{pmatrix}\mathrm{Decimal:}&-0.57735\dots \end{pmatrix}

Step-by-step explanation:

\cot \left(-600^{\circ \:}\right)

\gray{\mathrm{Use\:the\:following\:property:}\:\cot \left(-x\right)=-\cot \left(x\right)}

\gray{\cot \left(-600^{\circ \:}\right)=-\cot \left(600^{\circ \:}\right)}

=-\cot \left(600^{\circ \:}\right)

\black{\cot \left(600^{\circ \:}\right)}

\gray{\mathrm{Rewrite\:the\:angles\:for}\:\cot \left(600^{\circ \:}\right):}

$\displaystyle\gray{\cot \left(600^{\circ \:}\right)=\cot \left(\frac{9+1}{3}180^{\circ \:}\right)=\cot \left(\left(\frac{9}{3}+\frac{1}{3}\right)180^{\circ \:}\right)=\cot \left(180^{\circ \:}\cdot \:3+\frac{1}{3}180^{\circ \:}\right)}$

\displaystyle=\cot \left(180^{\circ \:}3+\frac{1}{3}180^{\circ \:}\right)

\gray{\mathrm{Use\:the\:periodiciity\:of\:}\cot :\quad \cot \left(x+180^{\circ \:}\cdot \:k\right)=\cot \left(x\right)}

\displaystyle\gray{\cot \left(180^{\circ \:}\cdot \:3+\frac{1}{3}180^{\circ \:}\right)=\cot \left(\frac{1}{3}180^{\circ \:}\right)}

\displaystyle=\cot \left(\frac{1}{3}180^{\circ \:}\right)

\gray{\mathrm{Simplify}}

=\cot \left(60^{\circ \:}\right)

=-\cot \left(60^{\circ \:}\right)

\displaystyle\gray{\mathrm{Use\:the\:following\:trivial\:identity}:\quad \cot \left(60^{\circ \:}\right)=\frac{\sqrt{3}}{3}}

\displaystyle=-\frac{\sqrt{3}}{3}

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