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Question: Question: If \({k_1} = \tan 27\theta - \tan \theta \) and \[{k_2} = \dfrac{{\sin \theta }}{{\cos 3\t...

Question: If k1=tan⁡27θ−tan⁡θ{k_1} = \tan 27\theta - \tan \theta and k2=sin⁡θcos⁡3θ+sin⁡3θcos⁡9θ+sin⁡9θcos⁡27θ{k_2} = \dfrac{{\sin \theta }}{{\cos 3\theta }} + \dfrac{{\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{\sin 9\theta }}{{\cos 27\theta }}, then
(A) k1=2k2{k_1} = 2{k_2}
(B) k1=k2+4{k_1} = {k_2} + 4
(C) k1=k2{k_1} = {k_2}
(D) None of these

Explanation

Solution

Compute the expressions tan⁡3θ−tan⁡θ\tan 3\theta - \tan \theta , tan⁡9θ−tan⁡3θ\tan 9\theta - \tan 3\theta and tan⁡27θ−tan⁡9θ\tan 27\theta - \tan 9\theta using the identities sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin (A - B) = \sin A\cos B - \cos A\sin Bandsin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A. Add the expressions tan⁡3θ−tan⁡θ\tan 3\theta - \tan \theta , tan⁡9θ−tan⁡3θ\tan 9\theta - \tan 3\theta and tan⁡27θ−tan⁡9θ\tan 27\theta - \tan 9\theta to get the value oftan⁡27θ−tan⁡θ\tan 27\theta - \tan \theta . Then substitute k1{k_1} and k2{k_2} to get the answer.

Complete step by step solution:
We are given k1=tan⁡27θ−tan⁡θ{k_1} = \tan 27\theta - \tan \theta and k2=sin⁡θcos⁡3θ+sin⁡3θcos⁡9θ+sin⁡9θcos⁡27θ{k_2} = \dfrac{{\sin \theta }}{{\cos 3\theta }} + \dfrac{{\sin 3\theta }}{{\cos 9\theta }} + \dfrac{{\sin 9\theta }}{{\cos 27\theta }}.
Now tan⁡3θ=sin⁡3θcos⁡3θ\tan 3\theta = \dfrac{{\sin 3\theta }}{{\cos 3\theta }}.
This gives us tan⁡3θ−tan⁡θ=sin⁡3θcos⁡3θ−sin⁡θcos⁡θ=sin⁡3θcos⁡θ−sin⁡θcos⁡3θcos⁡3θcos⁡θ\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⁡Acos⁡B−cos⁡Asin⁡B\sin (A - B) = \sin A\cos B - \cos A\sin Bandsin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A
Let A=3θA = 3\theta and B=θB = \theta
Therefore, we get tan⁡3θ−tan⁡θ=sin⁡(3θ−θ)cos⁡3θcos⁡θ=sin⁡2θcos⁡3θcos⁡θ=2sin⁡θcos⁡θcos⁡3θcos⁡θ=2sin⁡θcos⁡3θ\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⁡3θ−tan⁡θ=2sin⁡θcos⁡3θ......(1)\tan 3\theta - \tan \theta = \dfrac{{2\sin \theta }}{{\cos 3\theta }}......(1)
Using similar methods, we get
tan⁡9θ−tan⁡3θ=2sin⁡3θcos⁡9θ.....(2)\tan 9\theta - \tan 3\theta = \dfrac{{2\sin 3\theta }}{{\cos 9\theta }}.....(2)
tan⁡27θ−tan⁡9θ=2sin⁡9θcos⁡27θ....(3)\tan 27\theta - \tan 9\theta = \dfrac{{2\sin 9\theta }}{{\cos 27\theta }}....(3)
From (1), (2), and (3), we get
(tan⁡27θ−tan⁡9θ)+(tan⁡9θ−tan⁡3θ)+(tan⁡3θ−tan⁡θ)=2sin⁡9θcos⁡27θ+2sin⁡3θcos⁡9θ+2sin⁡θcos⁡3θ ⇒tan⁡27θ−tan⁡θ=2(sin⁡9θcos⁡27θ+sin⁡3θcos⁡9θ+sin⁡θcos⁡3θ)=2(sin⁡θcos⁡3θ+sin⁡3θcos⁡9θ+sin⁡9θcos⁡27θ)  (\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 k1=2k2{k_1} = 2{k_2}.
Hencek1=2k2{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 istan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}.