Question

The value of tan 3A – tan 2A – tan A is equal to:
(a) $\tan 3A\tan 2A\tan A$
(b) $-\tan 3A\tan 2A\tan A$
(c) $\tan A\tan 2A-\tan 2A\tan 3A-\tan 3A\tan A$
(d) None of these

Answer 1

Hint:Here, first we can write 3A = A + 2A and then apply tan on both the sides, we will get:
$\tan (3A)=\tan (A+2A)$.
Then we have to apply the trigonometric identity,
$\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
Now, with the help of this identity we can find the value of tan 3A – tan 2A – tan A.

Complete step-by-step answer:
Here, we have to find the value of tan 3A – tan 2A – tan A.
For that let us take,
3A = A + 2A
Next, by taking tan on both the sides,
$\Rightarrow \tan 3A=\tan (A+2A)$
We know by the trigonometric identity that,
$\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
Here, we have A in place of A and 2A in place of B and by applying this identity,
$\Rightarrow \tan 3A=\dfrac{\tan A+\tan 2A}{1-\tan A\tan 2A}$
Next, by cross multiplication,
$\Rightarrow \tan 3A(1-\tan A\tan 2A)=\tan A+\tan 2A$
Next, by multiplying with tan 3A we get,
$\begin{aligned} & \Rightarrow \tan 3A\times 1-\tan 3A\times \tan A\tan 2A=\tan A+\tan 2A \\\ & \Rightarrow \tan 3A-\tan 3A\tan A\tan 2A=\tan A+\tan 2A \\\ \end{aligned}$
Now, by taking $\tan A+\tan 2A$ to the left side it becomes $-\tan A-\tan 2A$ and $-\tan 3A\tan A\tan 2A$ to the right side we get $\tan 3A\tan A\tan 2A$,
$\Rightarrow \tan 3A-\tan 2A-\tan A=\tan 3A\tan A\tan 2A$
Hence, we can say that the value of $\tan 3A-\tan 2A-\tan A=\tan 3A\tan 2A\tan A$.
Therefore, the correct answer for this question is option (a).

Note: Here, we can also solve this problem by taking $\tan 3A-\tan 2A-\tan A=k$ and then taking $–\tan 2A – tan A$ to the right we get,
$\tan 3A=k+\tan 2A+\tan A$
Then apply the identity,
$\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ and find the value of k we will get required answer.