Question

If $\tan 2\theta .\tan 4\theta = 1$ then find the value of $\tan 3\theta$ .
A) $\dfrac{1}{2}$
B) 2
C) 0
D) 1

Answer 1

This is a trigonometric question. To solve the above problem first we will take one of the terms of the above equation to the opposite side and convert it into cot form followed by $\tan (90^\circ - \theta )$ form. After that we will compare the angles of both to get the value of $\theta$ and putting that value in $\tan 3\theta$ we will get the answer.

Complete step-by-step answer:
It is given in the question,
$\tan 2\theta .\tan 4\theta = 1$
Taking $\tan 4\theta$ to the right hand side of above equation we get,
$\tan 2\theta = \dfrac{1}{{\tan 4\theta }}$
We know that the formula of $\dfrac{1}{{\tan \theta }} = \cot \theta$ , Hence the above equation becomes
$\tan 2\theta = \cot 4\theta$
We know that the formula $\cot \theta = \tan (90^\circ - \theta )$ , Hence the above equation becomes
$\tan 2\theta = \tan (90^\circ - 4\theta )$
From the above equation we get that,
$2\theta = 90^\circ - 4\theta$
Taking $4\theta$ to the left hand side we get,
$2\theta + 4\theta = 90^\circ$
Simplifying the above equation we get,
$6\theta = 90^\circ$
$\Rightarrow \theta = \dfrac{{90^\circ }}{6} = 15^\circ$
$\therefore$ $3\theta = 3 \times 15^\circ = 45^\circ$
Hence, $\tan 3\theta = \tan 45^\circ = 1$

Option D is correct.

Note: There is an alternative method.
We know that if $\tan A.\tan B = 1$ , then $A + B = 90^\circ$
We are given in the question that,
$\tan 2\theta .\tan 4\theta = 1$
Hence, $(2\theta + 4\theta ) = 90^\circ$
Simplifying it we get,
$6\theta = 90^\circ$
$\Rightarrow \theta = \dfrac{{90^\circ }}{6} = 15^\circ$
$\therefore$ $3\theta = 3 \times 15^\circ = 45^\circ$
Hence, $\tan 3\theta = \tan 45^\circ = 1$ .
If you are getting confuse about the formula if $\tan A.\tan B = 1$ , then $A + B = 90^\circ$
Then we can prove it backward.
Let $A + B = 90^\circ$
Taking tan on both side we get,
$\tan (A + B) = \tan 90^\circ$
Putting the formula we get,
$\dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}} = \tan 90^\circ$
Putting the value of $\tan 90^\circ$ we get,
$\dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}} = \dfrac{1}{0}$ (As actual value of $\tan 90^\circ = \dfrac{1}{0}$ )
By cross multiplication we get,
$1 - \tan A.\tan B = 0$
$\tan A.\tan B = 1$
Hence proved.
Trigonometry is the type of mathematics that deals with the relationship between the sides and the angles of triangles.