# What is the value of $\tan {{\text{5}}^0}\tan 2{{\text{5}}^0}\tan {30^0}\tan 6{{\text{5}}^0}\tan

8{{\text{5}}^0} = $

$

\left( a \right){\text{ 1}} \\

\left( b \right){\text{ }}\dfrac{1}{2} \\

\left( c \right){\text{ }}\sqrt 3 \\

\left( d \right){\text{ }}\dfrac{1}{{\sqrt 3 }} \\

$

Last updated date: 27th Mar 2023

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Answer

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309.9k+ views

Hint- Use the basic trigonometric identity of $\tan (90 - \theta ) = \cot \theta $

Now we have to find the value of $\tan {{\text{5}}^0}\tan 2{{\text{5}}^0}\tan {30^0}\tan

6{{\text{5}}^0}\tan 8{{\text{5}}^0}$

Using $\tan {30^0} = \dfrac{1}{{\sqrt 3 }}$above we get

\[{\text{tan}}{{\text{5}}^0}{\text{tan2}}{{\text{5}}^0}\left( {\dfrac{1}{{\sqrt 3 }}}

\right){\text{tan6}}{{\text{5}}^0}{\text{tan8}}{{\text{5}}^0}\]

Now we can write \[{\text{tan6}}{{\text{5}}^0}\]as \[{\text{tan}}\left( {90 - 25} \right)\]and similar concept we will to \[{\text{tan8}}{{\text{5}}^0}\]

Thus we get

\[{\text{tan}}{{\text{5}}^0}{\text{tan2}}{{\text{5}}^0}\left( {\dfrac{1}{{\sqrt 3 }}} \right){\text{tan}}\left(

{90 - 25} \right){\text{tan}}\left( {90 - 5} \right)\]

Using the concept that $\tan (90 - \theta ) = \cot \theta $ we can rewrite the above as

\[{\text{tan}}{{\text{5}}^0}{\text{tan2}}{{\text{5}}^0}\left( {\dfrac{1}{{\sqrt 3 }}}

\right){\text{cot2}}{{\text{5}}^0}\cot {5^0}\]

As $\tan \theta = \dfrac{1}{{\cot \theta }}$

The above equation is simplified to \[\left( {\dfrac{1}{{\sqrt 3 }}} \right)\]

So option (d) is the right answer.

Note- The key concept that we need to recall every time we solve such type of problem is that always try

and convert one angle into other by subtracting or even sometimes adding it with the number that can

help changing the trigonometric term in order to cancel them with other terms to reach to the

simplified answer.

Now we have to find the value of $\tan {{\text{5}}^0}\tan 2{{\text{5}}^0}\tan {30^0}\tan

6{{\text{5}}^0}\tan 8{{\text{5}}^0}$

Using $\tan {30^0} = \dfrac{1}{{\sqrt 3 }}$above we get

\[{\text{tan}}{{\text{5}}^0}{\text{tan2}}{{\text{5}}^0}\left( {\dfrac{1}{{\sqrt 3 }}}

\right){\text{tan6}}{{\text{5}}^0}{\text{tan8}}{{\text{5}}^0}\]

Now we can write \[{\text{tan6}}{{\text{5}}^0}\]as \[{\text{tan}}\left( {90 - 25} \right)\]and similar concept we will to \[{\text{tan8}}{{\text{5}}^0}\]

Thus we get

\[{\text{tan}}{{\text{5}}^0}{\text{tan2}}{{\text{5}}^0}\left( {\dfrac{1}{{\sqrt 3 }}} \right){\text{tan}}\left(

{90 - 25} \right){\text{tan}}\left( {90 - 5} \right)\]

Using the concept that $\tan (90 - \theta ) = \cot \theta $ we can rewrite the above as

\[{\text{tan}}{{\text{5}}^0}{\text{tan2}}{{\text{5}}^0}\left( {\dfrac{1}{{\sqrt 3 }}}

\right){\text{cot2}}{{\text{5}}^0}\cot {5^0}\]

As $\tan \theta = \dfrac{1}{{\cot \theta }}$

The above equation is simplified to \[\left( {\dfrac{1}{{\sqrt 3 }}} \right)\]

So option (d) is the right answer.

Note- The key concept that we need to recall every time we solve such type of problem is that always try

and convert one angle into other by subtracting or even sometimes adding it with the number that can

help changing the trigonometric term in order to cancel them with other terms to reach to the

simplified answer.

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