Question

prove that tan6degree.tan21degree.tan30degree.tan69degree.tan84degree=1/root3

Answer 1

Answer:

Step-by-step explanation:

tan(90 degreee-84 degreee)=cot6 degree

tan(90-21) =cot 69 degree

so ,cot 6 degree.cot 69 degree.tan 30 degree.tan 69 degree .tan84 degree

so tan 69 and cot 69 will be canceled out similarly for tan 6 degree. now tan 30 =1/root 3

so if everything is canceled out except tan 30 then the answer is =

1/root 3

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Answer 2

$</p><p> \sf \rightarrow \quad tan \: {6}^{ \circ} \cdot tan \: {21}^{ \circ} \cdot tan \: {30}^{ \circ} \cdot tan \: {69}^{ \circ} \cdot tan \: {84}^{ \circ} \\ \\ \sf \rightarrow \quad cot \: (90 - 6) \cdot tan \: {30}^{ \circ} \cdot cot \: (90 - 21) \cdot tan \: {69}^{ \circ} \cdot tan \: {84}^{ \circ} \\ \\ \sf \quad tan \: (90 - \theta) = cot \: \theta \\ \\ \sf \rightarrow \quad \dfrac{1}{ \cancel{tan \: {84}^{ \circ} }} \cdot \dfrac{1}{ \cancel{tan \: {69}^{ \circ}} } \cdot tan \: {30}^{ \circ} \cdot \cancel{tan \: {69}^{ \circ} } \cdot \cancel{tan \: {84}^{ \circ} } \\ \\ \sf \quad cot \: \theta = \dfrac{1}{tan \: \theta} \\ \\ \sf \rightarrow \quad tan \: {30}^{ \circ} \\ \\ \sf \rightarrow \quad \frac{1}{ \sqrt{3} } \qquad \bigg( \: \because \: tan \: {30}^{ \circ} = \frac{1}{ \sqrt{3} } \: \bigg)</p><p>$