Question

If we have a trigonometric expression \[\dfrac{1}{{\tan 3A - \tan A}} - \dfrac{1}{{\cot 3A - \cot A}} = k\cot 2A\] , then $k$ is equal to
(A) $4$
(B) $3$
(C) $2$
(D) $1$

Answer 1

Hint: From the property of summation and subtraction we write that $2A = 3A - A$ and then we apply $\tan $ on both the sides, we will get $\tan 2A = \tan (3A - A) = \dfrac{{\tan 3A - \tan A}}{{1 + \tan A \times \tan 3A}}$ , now with the help of this identity we can find the value of the required variable $k$ . At first we use the properties of trigonometry that is $\tan A = \dfrac{1}{{\cot A}}$ i.e., $\cot A = \dfrac{1}{{\tan A}}$ . After that we use the above formula and we find the correct option.

Complete step-by-step solution:
The given function \[\dfrac{1}{{\tan 3A - \tan A}} - \dfrac{1}{{\cot 3A - \cot A}} = k\cot 2A\]
Now we use the properties of trigonometry , that is $\cot A = \dfrac{1}{{\cot A}}$ .
Therefore we have
\[\dfrac{1}{{\tan 3A - \tan A}} - \dfrac{1}{{\dfrac{1}{{\tan 3A}} - \dfrac{1}{{\tan A}}}} = k\cot 2A\]
Simplifying the above equation and we get
$ \Rightarrow \dfrac{1}{{\tan 3A - \tan A}} - \dfrac{1}{{\dfrac{{\tan A - \tan 3A}}{{\tan 3A \times \tan A}}}} = k\cot 2A$
Here we use the property of division $\dfrac{1}{{\dfrac{A}{B}}} = \dfrac{B}{A}$ in the above equation and we get
$ \Rightarrow \dfrac{1}{{\tan 3A - \tan A}} - \dfrac{{\tan 3A \times \tan A}}{{\tan A - \tan 3A}} = k\cot 2A$
We know that $A - B = - (B - A)$
Use this property in above equation and we get
$ \Rightarrow \dfrac{1}{{\tan 3A - \tan A}} + \dfrac{{\tan 3A \times \tan A}}{{\tan 3A - \tan A}} = k\cot 2A$
Now we calculate and simplifying the above equation and we get
$ \Rightarrow \dfrac{{1 - \tan 3A \times \tan A}}{{\tan 3A - \tan A}} = k\cot 2A$
$ \Rightarrow \dfrac{1}{{\dfrac{{\tan 3A - \tan A}}{{1 - \tan 3A \times \tan A}}}} = k\cot 2A$
Now we use the formula of $\tan $ i.e., $\tan (3A - A) = \dfrac{{\tan 3A - \tan A}}{{1 + \tan A \times \tan 3A}}$ , and we get
$ \Rightarrow \dfrac{1}{{\tan (3A - A)}} = k\cot 2A$
$ \Rightarrow \dfrac{1}{{\tan 2A}} = k\cot 2A$
$ \Rightarrow \cot 2A = k\cot 2A$
Cancelling the similar term and we get
$ \Rightarrow 1 = k$
i.e., $k = 1$
Therefore the value of $k$ is $1$
Option (D) is correct.

Note: We can also solve the given problem by using other properties and formulas of trigonometry. We use the property of trigonometry that is $\tan A = \dfrac{{\sin A}}{{\cos A}}$ and $\cot A = \dfrac{{\cos A}}{{\sin A}}$ and also we use use the formulas of trigonometry , that is $\sin (3A - A) = \sin 3A\cos A - \cos 3A\sin A$ and $\cos (3A - A) = \cos 3A\cos A + \sin 3A\sin A$ .
At first we use the property and we get
\[\dfrac{1}{{\dfrac{{\sin 3A}}{{\cos 3A}} - \dfrac{{\sin A}}{{\cos A}}}} - \dfrac{1}{{\dfrac{{\cos 3A}}{{\sin 3A}} - \dfrac{{\cos A}}{{\sin A}}}} = k\cot 2A\]
Now simply this equation and we get
\[ \Rightarrow \dfrac{1}{{\dfrac{{\sin 3A\cos A - \cos 3A\sin A}}{{\cos 3A \times \cos A}}}} - \dfrac{1}{{\dfrac{{\cos 3A\sin A - \sin 3A\cos A}}{{\sin 3A \times \sin A}}}} = k\cot 2A\]
\[ \Rightarrow \dfrac{{\cos 3A \times \cos A}}{{\sin 3A\cos A - \cos 3A\sin A}} + \dfrac{{\sin 3A \times \sin A}}{{\sin 3A\cos A - \cos 3A\sin A}} = k\cot 2A\]
Now we use the formula of sin of trigonometry and we get
\[ \Rightarrow \dfrac{{\cos 3A \times \cos A}}{{\sin (3A - A)}} + \dfrac{{\sin 3A \times \sin A}}{{\sin (3A - A)}} = k\cot 2A\]
\[ \Rightarrow \dfrac{{\cos 3A \times \cos A}}{{\sin 2A}} + \dfrac{{\sin 3A \times \sin A}}{{\sin 2A}} = k\cot 2A\]
Again we simplify this equation and we get
\[ \Rightarrow \dfrac{{\cos 3A \times \cos A + \sin 3A \times \sin A}}{{\sin 2A}} = k\cot 2A\]
Use the formula of cos of trigonometry and we get
\[ \Rightarrow \dfrac{{\cos (3A - A)}}{{\sin 2A}} = k\cot 2A\]
\[ \Rightarrow \dfrac{{\cos 2A}}{{\sin 2A}} = k\cot 2A\]
$ \Rightarrow \cot 2A = k\cot 2A$
Canceling the similar term and we get
$ \Rightarrow 1 = k$
i.e., $k = 1$
Therefore, Option (D) is correct.