Question

Prove the following $\tan {36^0}\tan {17^0}\tan {54^0}\tan {73^0} = 1$

Answer 1

Hint: In order to solve the problem try to break the angle as a sum of other angle with multiple of\[{90^ \circ },{180^ \circ },{270^ \circ }\& {360^ \circ }\]. Try to use trigonometric identities to form such a pair whose product will be 1.

Complete step-by-step answer:

We have to prove $\tan {36^0}\tan {17^0}\tan {54^0}\tan {73^0} = 1$
Let us proceed with the LHS of the equation
$ \Rightarrow \tan {36^0}\tan {17^0}\tan {54^0}\tan {73^0}$
Let us make the pair of angles whose sum is ${90^0}$
$ \Rightarrow \left( {\tan {{36}^0}\tan {{54}^0}} \right)\left( {\tan {{17}^0}\tan {{73}^0}} \right)$ -----(A)
As we know the algebraic relation connecting $\tan ,\cot {\text{ and }}{90^0}$ which is given as
$
   \Rightarrow \tan \theta = \cot \left( {{{90}^0} - \theta } \right) \\
  \because \cot \theta = \dfrac{1}{{\tan \theta }} \\
  \therefore \tan \theta = \cot \left( {{{90}^0} - \theta } \right) = \dfrac{1}{{\tan \left( {{{90}^0} - \theta } \right)}} \\
 $
Using the above relation in one out of the two angles in each bracket.
For $\tan \left( {{{54}^0}} \right)$
$
   \Rightarrow \tan \left( {{{54}^0}} \right) = \cot \left( {{{90}^0} - {{54}^0}} \right) = \dfrac{1}{{\tan \left( {{{90}^0} - {{54}^0}} \right)}} \\
   \Rightarrow \tan \left( {{{54}^0}} \right) = \dfrac{1}{{\tan \left( {{{36}^0}} \right)}}..............(1) \\
 $
For $\tan \left( {{{73}^0}} \right)$
$
   \Rightarrow \tan \left( {{{73}^0}} \right) = \cot \left( {{{90}^0} - {{73}^0}} \right) = \dfrac{1}{{\tan \left( {{{90}^0} - {{73}^0}} \right)}} \\
   \Rightarrow \tan \left( {{{54}^0}} \right) = \dfrac{1}{{\tan \left( {{{17}^0}} \right)}}..............(2) \\
 $
Now using the results of equation (1) and equation (2) in problem equation (A)
\[
   \Rightarrow \left( {\tan {{36}^0}\tan {{54}^0}} \right)\left( {\tan {{17}^0}\tan {{73}^0}} \right) \\
   = \left( {\tan {{36}^0} \times \dfrac{1}{{\tan {{36}^0}}}} \right)\left( {\tan {{17}^0} \times \dfrac{1}{{\tan {{17}^0}}}} \right){\text{ }}\left[ {\because \tan {{73}^0} = \dfrac{1}{{\tan {{17}^0}}}\& \tan {{54}^0} = \dfrac{1}{{\tan {{36}^0}}}} \right] \\
   = 1 \times 1 \\
   = 1 \\
 \]
Since we have got the solution of LHS = 1 =RHS, Hence the given equation is proved.

Note: The following problem cannot be solved by putting in the values of each of the terms, as it is not possible to remember trigonometric values for such angles. It is easier to solve the problem by rearranging the terms and bringing it in simpler terms. Also some of the common trigonometric identities must be remembered.