Question

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Answer 1

Answer:

Hope it will helpful..

Answer 2

To prove :

$\sf \tan(30^{\circ} ) \: = \: \dfrac{ \tan(60 ^{\circ} ) - \tan(30 ^{\circ} ) }{1 + \tan(30 ^{\circ} ) \times \tan(30 ^{\circ}) }$

Trigonometric table :

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

{ User using through app refer to attached image for the same }

Solution :

LHS ➝

$\sf \tan( {30}^{ \circ} ) \: = \: \dfrac{1}{ \sqrt{ 3} }$

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RHS ➝

$\sf \implies \dfrac{ \sqrt{3} - \dfrac{1}{ \sqrt{3} } }{1 + ( \sqrt{3} \times \dfrac{1}{ \sqrt{3} }) } \\ \\ \sf \: takimg \: lcm \: we \: get \\ \\ \sf \implies \dfrac{ \dfrac{( \sqrt{3} \times \sqrt{3}) - (1 \times 1) }{ \sqrt{3} } }{1 + 1} \\ \\ \sf \implies \dfrac{ \dfrac{3 - 1}{ \sqrt{3} } }{2} \\ \\ \sf \implies \dfrac{2}{ \sqrt{3} } \times \dfrac{1}{2} \\ \\ \sf \implies \dfrac{1}{ \sqrt{3} }$

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➝ LHS = (1/√3) = RHS

➝ LHS = RHS

Hence proved!