Question

# What is the value of $\tan {{\text{5}}^0}\tan 2{{\text{5}}^0}\tan {30^0}\tan 6{{\text{5}}^0}\tanÂ 8{{\text{5}}^0} =$$Â Â \left( a \right){\text{ 1}} \\Â Â \left( b \right){\text{ }}\dfrac{1}{2} \\Â Â \left( c \right){\text{ }}\sqrt 3 \\Â Â \left( d \right){\text{ }}\dfrac{1}{{\sqrt 3 }} \\ Â$

Hint- Use the basic trigonometric identity of $\tan (90 - \theta ) = \cot \theta$
Now we have to find the value of $\tan {{\text{5}}^0}\tan 2{{\text{5}}^0}\tan {30^0}\tan Â 6{{\text{5}}^0}\tan 8{{\text{5}}^0}$
Using $\tan {30^0} = \dfrac{1}{{\sqrt 3 }}$above we get
${\text{tan}}{{\text{5}}^0}{\text{tan2}}{{\text{5}}^0}\left( {\dfrac{1}{{\sqrt 3 }}} Â \right){\text{tan6}}{{\text{5}}^0}{\text{tan8}}{{\text{5}}^0}$
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
${\text{tan}}{{\text{5}}^0}{\text{tan2}}{{\text{5}}^0}\left( {\dfrac{1}{{\sqrt 3 }}} \right){\text{tan}}\left( Â {90 - 25} \right){\text{tan}}\left( {90 - 5} \right)$
Using the concept that $\tan (90 - \theta ) = \cot \theta$ we can rewrite the above as
${\text{tan}}{{\text{5}}^0}{\text{tan2}}{{\text{5}}^0}\left( {\dfrac{1}{{\sqrt 3 }}} Â \right){\text{cot2}}{{\text{5}}^0}\cot {5^0}$
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
Â and convert one angle into other by subtracting or even sometimes adding it with the number that can
Â help changing the trigonometric term in order to cancel them with other terms to reach to the