Question

If ${k_1} = \tan 27\theta - \tan \theta $ and ${k_2} = \dfrac{{\sin \theta }}{{\cos 3\theta }} + \dfrac{{\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{\sin 9\theta }}{{\cos 27\theta }}$ , then
A) ${k_1} = 2{k_2}$
B) ${k_1} = {k_2} + 4$
C) ${k_1} = {k_2}$
D) None of these

Answer 1

Hint:
It is given in the question that ${k_1} = \tan 27\theta - \tan \theta $ and ${k_2} = \dfrac{{\sin \theta }}{{\cos 3\theta }} + \dfrac{{\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{\sin 9\theta }}{{\cos 27\theta }}$
Firstly, we will find the value of $\tan 3\theta - \tan \theta $ , $\tan 9\theta - \tan 3\theta $ , $\tan 27\theta - \tan 9\theta $ using formula $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ and $\sin 2A = 2\sin A\cos A$ .
Then after, we will add the values of $\tan 3\theta - \tan \theta $ , $\tan 9\theta - \tan 3\theta $ , $\tan 27\theta - \tan 9\theta $ to get ${k_1} = \tan 27\theta - \tan \theta $ and ${k_2} = \dfrac{{\sin \theta }}{{\cos 3\theta }} + \dfrac{{\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{\sin 9\theta }}{{\cos 27\theta }}$ .
Finally, solving further we will get the required answer.

Complete step by step solution:
It is given in the question that ${k_1} = \tan 27\theta - \tan \theta $ and ${k_2} = \dfrac{{\sin \theta }}{{\cos 3\theta }} + \dfrac{{\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{\sin 9\theta }}{{\cos 27\theta }}$
Now, first $\tan 3\theta = \dfrac{{\sin 3\theta }}{{\cos 3\theta }}$ .
 $\therefore \tan 3\theta - \tan \theta = \dfrac{{\sin 3\theta }}{{\cos 3\theta }} - \dfrac{{\sin \theta }}{{\cos \theta }}$
 $\therefore \tan 3\theta - \tan \theta = \dfrac{{\sin 3\theta \cos \theta - \sin \theta \cos 3\theta }}{{\cos 3\theta \cos \theta }}$
Since, we know that $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ and $\sin 2A = 2\sin A\cos A$ .
Now, let $A = 3\theta $ and $B = \theta $ .
Therefore, by applying formula in the above equation, we get,
 $\therefore \tan 3\theta - \tan \theta = \dfrac{{\sin \left( {3\theta - \theta } \right)}}{{\cos 3\theta \cos \theta }}$
 \[\therefore \tan 3\theta - \tan \theta = \dfrac{{\sin 2\theta }}{{\cos 3\theta \cos \theta }}\]
Since, $\sin 2\theta = 2\sin \theta \cos \theta $
 \[\therefore \tan 3\theta - \tan \theta = \dfrac{{2\sin \theta \cos \theta }}{{\cos 3\theta \cos \theta }}\]
 \[\therefore \tan 3\theta - \tan \theta = \dfrac{{2\sin \theta }}{{\cos 3\theta }}\] (I)
Similarly, using above method we will find the value of $\tan 9\theta - \tan 3\theta $
 $\therefore \tan 9\theta - \tan 3\theta = \dfrac{{\sin 9\theta }}{{\cos 9\theta }} - \dfrac{{\sin 3\theta }}{{\cos 3\theta }}$
 $\therefore \tan 9\theta - \tan 3\theta = \dfrac{{\sin 9\theta \cos 3\theta - \sin 3\theta \cos 9\theta }}{{\cos 9\theta \cos 3\theta }}$
 $\therefore \tan 9\theta - \tan 3\theta = \dfrac{{\sin \left( {9\theta - 3\theta } \right)}}{{\cos 9\theta \cos 3\theta }}$
 \[\therefore \tan 9\theta - \tan 3\theta = \dfrac{{\sin 6\theta }}{{\cos 9\theta \cos 3\theta }}\]
 \[\therefore \tan 9\theta - \tan 3\theta = \dfrac{{2\sin 3\theta \cos 3\theta }}{{\cos 9\theta \cos 3\theta }}\]
 \[\therefore \tan 9\theta - \tan 3\theta = \dfrac{{2\sin 3\theta }}{{\cos 9\theta }}\] (II)
Similarly, using above method we will find the value of $\tan 27\theta - \tan 9\theta $
 $\therefore \tan 27\theta - \tan 9\theta = \dfrac{{\sin 27\theta }}{{\cos 27\theta }} - \dfrac{{\sin 9\theta }}{{\cos 9\theta }}$
 $\therefore \tan 27\theta - \tan 9\theta = \dfrac{{\sin 27\theta \cos 9\theta - \sin 9\theta \cos 27\theta }}{{\cos 27\theta \cos 9\theta }}$
 $\therefore \tan 27\theta - \tan 9\theta = \dfrac{{\sin \left( {27\theta - 9\theta } \right)}}{{\cos 27\theta \cos 9\theta }}$
 \[\therefore \tan 27\theta - \tan 9\theta = \dfrac{{\sin 18\theta }}{{\cos 27\theta \cos 9\theta }}\]
 \[\therefore \tan 27\theta - \tan 9\theta = \dfrac{{2\sin 9\theta \cos 9\theta }}{{\cos 27\theta \cos 9\theta }}\]
 \[\therefore \tan 27\theta - \tan 9\theta = \dfrac{{2\sin 9\theta }}{{\cos 27\theta }}\] (III)
Now, add equation (I), (II) and (III), we get,
  $\therefore \left( {\tan 27\theta - \tan 9\theta } \right) + \left( {\tan 9\theta - \tan 3\theta } \right) + \left( {\tan 3\theta - \tan \theta } \right) = \dfrac{{2\sin 9\theta }}{{\cos 27\theta }} + \dfrac{{2\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{2\sin \theta }}{{\cos 3\theta }}$
 $\therefore \left( {\tan 27\theta - \tan \theta } \right) = 2\left( {\dfrac{{\sin 9\theta }}{{\cos 27\theta }} + \dfrac{{\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{\sin \theta }}{{\cos 3\theta }}} \right)$
 $\therefore \left( {\tan 27\theta - \tan \theta } \right) = 2\left( {\dfrac{{\sin \theta }}{{\cos 3\theta }} + \dfrac{{\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{\sin 9\theta }}{{\cos 27\theta }}} \right)$
Since, we have given the question that ${k_1} = \tan 27\theta - \tan \theta $ and ${k_2} = \dfrac{{\sin \theta }}{{\cos 3\theta }} + \dfrac{{\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{\sin 9\theta }}{{\cos 27\theta }}$
 $\therefore {k_1} = 2{k_2}$

Hence, option (A) is correct.

Note:
As $\sin ,\cos $ and $\tan $ functions are the most commonly used trigonometry ratio.
Some properties containing functions of $\sin ,\cos $ and $\tan $ are as follows:
 $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
 ${\sin ^2}\theta + {\cos ^2}\theta = 1$
 $\sin 2\theta = 2\sin \theta \cos \theta $
 $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $
 $\tan 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$