Question

{sec 180°+tan150°}÷{cosec90°-cot120°}=2-√3​

Answer 1

Answer:

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${ \tt{(sec180 + tan150) \div (cosec90 - cot120) = \frac{2 }{\sqrt{3} } }}$

Solution:-

${ \implies{ \tt{sec180 = sec(180 - 0) = sec0 = 0}}}$

${ \implies{ \sf{tan150 = tan(180 - 30) = tan30 = \frac{1}{ \sqrt{3} } }}}$

${ \implies{ \sf{cosec90 = 1}}}$

${ \implies{ \sf{cot120 = cot(90 + 30) = sin30 = \frac{1}{2} }}}$

Therefore:-

Sec180° = 0

Tan 150° = 1/√3

Cosec90° = 1

Cot120°=1/2

${ \sf{(sec180 + tan150) \div (cosec90 - cot120) }}$

${ \implies{ \sf({0 + \frac{1}{ \sqrt{3} } ) \div 1 - \frac{1}{2} = \frac{2 - \sqrt{3} }{2 \sqrt{3} } }}}$

${ \implies{ \sf{ \frac{1}{ \sqrt{3} } \div \frac{1}{2} = \frac{2 - \sqrt{3} }{2 \sqrt{3} } }}}$

${ \implies{ \sf{ \frac{2 }{\sqrt{3} } = \frac{2}{\sqrt{3} } }}}$

Hence proved ✔

Related information:-

Sin (90-θ)=cosθ

Cos(90-θ)=sinθ

Tan(90-θ)=cotθ

Cot(90-θ)=tanθ

Sec(90-θ)=cscθ

Csc(90-θ)=secθ

________________

Sin (180-θ)=sinθ

Cos(180-θ)=cosθ

Tan(180-θ)=tanθ

Cot(180-θ)=cotθ

Sec(180-θ)=secθ

Csc(180-θ)=cscθ

Answer 2

$\to\ \sf{\bf{\dfrac{sec\ 180+tan\ 150}{csc\ 90-cot\ 120}}}$

$\to \sf \dfrac{\frac{1}{cos\ 180}+tan(90+60)}{\frac{1}{sin\ 90}-cot(90+30)}$

$\to \sf \dfrac{\frac{1}{-1}+(-cot\ 60)}{\frac{1}{1}-(-tan\ 30)}$

$\to\ \sf \dfrac{-1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$

$\to \sf \dfrac{\frac{-\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}$

$\to \sf \dfrac{-(\sqrt{3}+1)}{\sqrt{3}+1}$

$\to\ \sf -1$