Question

Prove that:
$\tan {36^ \circ } + \tan {9^ \circ } + \tan {36^ \circ }\tan {9^ \circ } = 1$

Answer 1

Here first we will use the known value of $\tan {45^ \circ }$ and then apply the trigonometric identity and solve the equation so formed to get the desired answer.

$$\tan {45^ \circ } = 1 \\\ \tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} \\\$$

Complete step-by-step answer:
We have to prove that:
$\tan {36^ \circ } + \tan {9^ \circ } + \tan {36^ \circ }\tan {9^ \circ } = 1$
Now we know that:-
$\tan {45^ \circ } = 1$…………………………………….(1)
Also,
${45^ \circ } = {36^ \circ } + {9^ \circ }$
Hence substituting the value in the equation 1 we get:-
$\tan \left( {{{36}^ \circ } + {9^ \circ }} \right) = 1$
Now applying the following identity:-
$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
We get:-

$$\tan \left( {{{36}^ \circ } + {9^ \circ }} \right) = \dfrac{{\tan {{36}^ \circ } + \tan {9^ \circ }}}{{1 - \tan {{36}^ \circ }\tan {9^ \circ }}} \\\ \Rightarrow \tan {45^ \circ } = \dfrac{{\tan {{36}^ \circ } + \tan {9^ \circ }}}{{1 - \tan {{36}^ \circ }\tan {9^ \circ }}} \\\$$

Now putting value from equation 1 we get:-
$1 = \dfrac{{\tan {{36}^ \circ } + \tan {9^ \circ }}}{{1 - \tan {{36}^ \circ }\tan {9^ \circ }}}$
Now on cross multiplying we get:-
$1 - \tan {36^ \circ }\tan {9^ \circ } = \tan {36^ \circ } + \tan {9^ \circ }$
Now simplifying it further we get:-
$\tan {36^ \circ } + \tan {9^ \circ } + \tan {36^ \circ }\tan {9^ \circ } = 1$
Hence proved.

Note: In such questions we have to form the already known identity to simplify the given expression and get the desired answer.
We cannot directly proceed in this question as we don’t know the values of $\tan {36^ \circ }$ and $\tan {9^ \circ }$.
Also, students should notice that on adding the given angles we get an angle whose value is already known. The identities used should be correct and appropriate to get the correct answer.