Question

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: Compute the expressions $\tan 3\theta - \tan \theta $, $\tan 9\theta - \tan 3\theta $ and $\tan 27\theta - \tan 9\theta $ using the identities $\sin (A - B) = \sin A\cos B - \cos A\sin B$and$\sin 2A = 2\sin A\cos A$. Add the expressions $\tan 3\theta - \tan \theta $, $\tan 9\theta - \tan 3\theta $ and $\tan 27\theta - \tan 9\theta $to get the value of$\tan 27\theta - \tan \theta $. Then substitute ${k_1}$ and ${k_2}$ to get the answer.


Complete step by step solution:
We are given ${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 $\tan 3\theta = \dfrac{{\sin 3\theta }}{{\cos 3\theta }}$.
This gives us $\tan 3\theta - \tan \theta = \dfrac{{\sin 3\theta }}{{\cos 3\theta }} - \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{\sin 3\theta \cos \theta - \sin \theta \cos 3\theta }}{{\cos 3\theta \cos \theta }}$
We know that $\sin (A - B) = \sin A\cos B - \cos A\sin B$and$\sin 2A = 2\sin A\cos A$
Let $A = 3\theta $and $B = \theta $
Therefore, we get $\tan 3\theta - \tan \theta = \dfrac{{\sin (3\theta - \theta )}}{{\cos 3\theta \cos \theta }} = \dfrac{{\sin 2\theta }}{{\cos 3\theta \cos \theta }} = \dfrac{{2\sin \theta \cos \theta }}{{\cos 3\theta \cos \theta }} = \dfrac{{2\sin \theta }}{{\cos 3\theta }}$
That is $\tan 3\theta - \tan \theta = \dfrac{{2\sin \theta }}{{\cos 3\theta }}......(1)$
Using similar methods, we get
$\tan 9\theta - \tan 3\theta = \dfrac{{2\sin 3\theta }}{{\cos 9\theta }}.....(2)$
$\tan 27\theta - \tan 9\theta = \dfrac{{2\sin 9\theta }}{{\cos 27\theta }}....(3)$
From (1), (2), and (3), we get
$
  (\tan 27\theta - \tan 9\theta ) + (\tan 9\theta - \tan 3\theta ) + (\tan 3\theta - \tan \theta ) = \dfrac{{2\sin 9\theta }}{{\cos 27\theta }} + \dfrac{{2\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{2\sin \theta }}{{\cos 3\theta }} \\
   \Rightarrow \tan 27\theta - \tan \theta = 2(\dfrac{{\sin 9\theta }}{{\cos 27\theta }} + \dfrac{{\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{\sin \theta }}{{\cos 3\theta }}) = 2(\dfrac{{\sin \theta }}{{\cos 3\theta }} + \dfrac{{\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{\sin 9\theta }}{{\cos 27\theta }}) \\
 $
Thus, using the given information we get \[{k_1} = 2{k_2}\].
Hence\[{k_1} = 2{k_2}\] is the correct answer.
Correct Option: (A)


Note: sine, cosine and tangent are the most commonly used trigonometric ratios. Therefore, knowing the identities which relate these ratios with each other helps solve many problems.
One of the most frequently used identities is$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$.