# Prove that \[\dfrac{1}{\sin {{10}^{o}}}-\dfrac{\sqrt{3}}{\cos {{10}^{o}}}=4\].

Last updated date: 16th Mar 2023

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Answer

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Hint: Use formula \[\sin 2\theta =2\sin \theta \cos \theta \]and express equation in form of

\[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\].

We have to prove that \[\dfrac{1}{\sin {{10}^{o}}}-\dfrac{\sqrt{3}}{\cos {{10}^{o}}}=4....\left( i \right)\]

Taking left hand side of equation \[\left( i \right)\]

\[\Rightarrow \dfrac{1}{\sin {{10}^{o}}}-\dfrac{\sqrt{3}}{\cos {{10}^{o}}}\]

\[=\dfrac{\cos {{10}^{o}}-\left( \sin {{10}^{o}} \right).\sqrt{3}}{\sin {{10}^{o}}.\cos {{10}^{o}}}\]

Now, we will multiply numerator and denominator by \[2\].

We get, \[\dfrac{2\left( \cos {{10}^{o}}-\sqrt{3}\sin {{10}^{o}} \right)}{2\sin {{10}^{o}}.\cos {{10}^{o}}}....\left( ii \right)\]

Now, we know that \[2\sin \theta \cos \theta =\sin 2\theta \]

Therefore, \[2\sin {{10}^{o}}.\cos {{10}^{o}}=\sin 2\left( {{10}^{o}} \right)\]\[=\sin {{20}^{o}}\]

We will put the value of \[2\sin {{10}^{o}}\cos {{10}^{o}}\]in equation \[\left( ii \right)\].

We get \[\dfrac{2\left( \cos {{10}^{o}}-\sqrt{3}\sin {{10}^{o}} \right)}{\sin {{20}^{o}}}\]

We will multiply and divide numerator by \[2\].

We get, \[\dfrac{2.2.\dfrac{1}{2}\left( \cos {{10}^{o}}-\sqrt{3}\sin {{10}^{o}} \right)}{\sin {{20}^{o}}}\]

By rearranging the equation and taking \[\dfrac{1}{2}\]inside the bracket.

We get, \[\dfrac{4\left( \dfrac{1}{2}\cos {{10}^{o}}-\dfrac{\sqrt{3}}{2}\sin {{10}^{o}} \right)}{\sin {{20}^{o}}}...\left( iii \right)\]

Now, we know that \[\sin {{30}^{o}}=\dfrac{1}{2}\]and \[\cos {{30}^{o}}=\dfrac{\sqrt{3}}{2}\].

We put these values in equation \[\left( iii \right)\].

We get \[\dfrac{4\left( \sin {{30}^{o}}\cos {{10}^{o}}-\cos {{30}^{o}}\sin {{10}^{o}} \right)}{\sin {{20}^{o}}}....\left( iv \right)\]

We know that \[\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)\]

Therefore, \[\sin {{30}^{o}}\cos {{10}^{o}}-\cos {{30}^{o}}\sin {{10}^{o}}=\sin \left( {{30}^{o}}-{{10}^{o}} \right)=\sin {{20}^{o}}\]

Putting these values in equation \[\left( iv \right)\]

We get, \[\dfrac{4\sin {{20}^{o}}}{\sin {{20}^{o}}}\]

\[=4\]\[=\]Right hand side or RHS

Therefore, \[LHS=RHS\][Hence Proved]

Note: Some students try to use \[\dfrac{1}{\sin {{10}^{o}}}=\operatorname{cosec}{{10}^{o}}\] and

\[\dfrac{1}{\cos {{10}^{o}}}=\sec {{10}^{o}}\]in question to get away with fractions, but this actually

creates confusion and makes solution lengthy.

\[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\].

We have to prove that \[\dfrac{1}{\sin {{10}^{o}}}-\dfrac{\sqrt{3}}{\cos {{10}^{o}}}=4....\left( i \right)\]

Taking left hand side of equation \[\left( i \right)\]

\[\Rightarrow \dfrac{1}{\sin {{10}^{o}}}-\dfrac{\sqrt{3}}{\cos {{10}^{o}}}\]

\[=\dfrac{\cos {{10}^{o}}-\left( \sin {{10}^{o}} \right).\sqrt{3}}{\sin {{10}^{o}}.\cos {{10}^{o}}}\]

Now, we will multiply numerator and denominator by \[2\].

We get, \[\dfrac{2\left( \cos {{10}^{o}}-\sqrt{3}\sin {{10}^{o}} \right)}{2\sin {{10}^{o}}.\cos {{10}^{o}}}....\left( ii \right)\]

Now, we know that \[2\sin \theta \cos \theta =\sin 2\theta \]

Therefore, \[2\sin {{10}^{o}}.\cos {{10}^{o}}=\sin 2\left( {{10}^{o}} \right)\]\[=\sin {{20}^{o}}\]

We will put the value of \[2\sin {{10}^{o}}\cos {{10}^{o}}\]in equation \[\left( ii \right)\].

We get \[\dfrac{2\left( \cos {{10}^{o}}-\sqrt{3}\sin {{10}^{o}} \right)}{\sin {{20}^{o}}}\]

We will multiply and divide numerator by \[2\].

We get, \[\dfrac{2.2.\dfrac{1}{2}\left( \cos {{10}^{o}}-\sqrt{3}\sin {{10}^{o}} \right)}{\sin {{20}^{o}}}\]

By rearranging the equation and taking \[\dfrac{1}{2}\]inside the bracket.

We get, \[\dfrac{4\left( \dfrac{1}{2}\cos {{10}^{o}}-\dfrac{\sqrt{3}}{2}\sin {{10}^{o}} \right)}{\sin {{20}^{o}}}...\left( iii \right)\]

Now, we know that \[\sin {{30}^{o}}=\dfrac{1}{2}\]and \[\cos {{30}^{o}}=\dfrac{\sqrt{3}}{2}\].

We put these values in equation \[\left( iii \right)\].

We get \[\dfrac{4\left( \sin {{30}^{o}}\cos {{10}^{o}}-\cos {{30}^{o}}\sin {{10}^{o}} \right)}{\sin {{20}^{o}}}....\left( iv \right)\]

We know that \[\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)\]

Therefore, \[\sin {{30}^{o}}\cos {{10}^{o}}-\cos {{30}^{o}}\sin {{10}^{o}}=\sin \left( {{30}^{o}}-{{10}^{o}} \right)=\sin {{20}^{o}}\]

Putting these values in equation \[\left( iv \right)\]

We get, \[\dfrac{4\sin {{20}^{o}}}{\sin {{20}^{o}}}\]

\[=4\]\[=\]Right hand side or RHS

Therefore, \[LHS=RHS\][Hence Proved]

Note: Some students try to use \[\dfrac{1}{\sin {{10}^{o}}}=\operatorname{cosec}{{10}^{o}}\] and

\[\dfrac{1}{\cos {{10}^{o}}}=\sec {{10}^{o}}\]in question to get away with fractions, but this actually

creates confusion and makes solution lengthy.

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