Question

If $\tan 7\theta .\tan 3\theta = 1$ , then find the value of $ - \theta $.
A. ${0^ \circ }$
B. ${9^ \circ }$
C. ${10^ \circ }$
D. ${18^ \circ }$

Answer 1

Hint: We will first express $\tan 7\theta .\tan 3\theta $ in the form of $\tan 10\theta $ and then find the value of $\tan 10\theta $ comes out as infinity. After that, we will equate $10\theta $ to $\dfrac{\pi }{2}$ and get the required answer.

Complete step-by-step answer:
Let us first get to know the formula of tan (a+b):
$\tan (a + b) = \dfrac{{\tan a + \tan b}}{{1 - \tan a.\tan b}}$.
Taking $a = 7\theta $ and $b = 3\theta $ in this formula, we will get:-
$\tan (7\theta + 3\theta ) = \dfrac{{\tan 7\theta + \tan 3\theta }}{{1 - \tan 7\theta .\tan 3\theta }}$
Simplifying it will result I as follows:-
$\tan 10\theta = \dfrac{{\tan 7\theta + \tan 3\theta }}{{1 - \tan 7\theta .\tan 3\theta }}$ …………(1)
We are given the question that $\tan 7\theta .\tan 3\theta = 1$.
Taking 1 from RHS to LHS and then multiplying the whole equation with -1, we will get:-
$1 - \tan 7\theta .\tan 3\theta = 0$
Putting this value in (1), we will get:-
$\tan 10\theta = \dfrac{{\tan 7\theta + \tan 3\theta }}{0}$
Since, the denominator on RHS is 0, hence it is undefined and we denote undefined by the term ‘infinity’.
Therefore, we have:- $\tan 10\theta = \infty $
We know that tangent is infinity at $\dfrac{\pi }{2}$.
Therefore, we now get:- $10\theta = \dfrac{\pi }{2}$.
Taking 10 from LHS to division in RHS, we will get:-
$\theta = \dfrac{\pi }{{2 \times 10}} = \dfrac{\pi }{{20}}$.
Now, let us convert the angle in degrees from radians.
We know that $\pi = {180^ \circ }$.
Therefore, $\dfrac{\pi }{{20}} = \dfrac{1}{{20}} \times {180^ \circ } = {9^ \circ }$.
We ignored negative because multiplication of two negative numbers is positive.

So, the correct answer is “Option B”.

Note: There is an alternate way to do the same question which has a bit more hassle and may also require a calculator. The way is, you can put in every option from the given options and see if they satisfy the equation. So, whichever ones do satisfy can be the possible answer of the question. But you cannot rely on that because we are sometimes not allowed to use calculators during exams.
The students must note that we could ignore negative because both the angle $7\theta $ and $2\theta $ were acute even after putting in the required values. And $\tan ( - \theta ) = - \tan \theta $. If some angles exceed the right angle, it will reach the third quadrant where the tangent is positive.