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

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

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

Complete step by step solution:
It is given in the question that 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, first tan⁡3θ=sin⁡3θcos⁡3θ\tan 3\theta = \dfrac{{\sin 3\theta }}{{\cos 3\theta }} .
∴tan⁡3θ−tan⁡θ=sin⁡3θcos⁡3θ−sin⁡θcos⁡θ\therefore \tan 3\theta - \tan \theta = \dfrac{{\sin 3\theta }}{{\cos 3\theta }} - \dfrac{{\sin \theta }}{{\cos \theta }}
∴tan⁡3θ−tan⁡θ=sin⁡3θcos⁡θ−sin⁡θcos⁡3θcos⁡3θcos⁡θ\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⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B and sin⁡2A=2sin⁡Acos⁡A\sin 2A = 2\sin A\cos A .
Now, let A=3θA = 3\theta and B=θB = \theta .
Therefore, by applying formula in the above equation, we get,
∴tan⁡3θ−tan⁡θ=sin⁡(3θ−θ)cos⁡3θcos⁡θ\therefore \tan 3\theta - \tan \theta = \dfrac{{\sin \left( {3\theta - \theta } \right)}}{{\cos 3\theta \cos \theta }}
∴tan⁡3θ−tan⁡θ=sin⁡2θcos⁡3θcos⁡θ\therefore \tan 3\theta - \tan \theta = \dfrac{{\sin 2\theta }}{{\cos 3\theta \cos \theta }}
Since, sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta
∴tan⁡3θ−tan⁡θ=2sin⁡θcos⁡θcos⁡3θcos⁡θ\therefore \tan 3\theta - \tan \theta = \dfrac{{2\sin \theta \cos \theta }}{{\cos 3\theta \cos \theta }}
∴tan⁡3θ−tan⁡θ=2sin⁡θcos⁡3θ\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⁡9θ−tan⁡3θ\tan 9\theta - \tan 3\theta
∴tan⁡9θ−tan⁡3θ=sin⁡9θcos⁡9θ−sin⁡3θcos⁡3θ\therefore \tan 9\theta - \tan 3\theta = \dfrac{{\sin 9\theta }}{{\cos 9\theta }} - \dfrac{{\sin 3\theta }}{{\cos 3\theta }}
∴tan⁡9θ−tan⁡3θ=sin⁡9θcos⁡3θ−sin⁡3θcos⁡9θcos⁡9θcos⁡3θ\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 }}
∴tan⁡9θ−tan⁡3θ=sin⁡(9θ−3θ)cos⁡9θcos⁡3θ\therefore \tan 9\theta - \tan 3\theta = \dfrac{{\sin \left( {9\theta - 3\theta } \right)}}{{\cos 9\theta \cos 3\theta }}
∴tan⁡9θ−tan⁡3θ=sin⁡6θcos⁡9θcos⁡3θ\therefore \tan 9\theta - \tan 3\theta = \dfrac{{\sin 6\theta }}{{\cos 9\theta \cos 3\theta }}
∴tan⁡9θ−tan⁡3θ=2sin⁡3θcos⁡3θcos⁡9θcos⁡3θ\therefore \tan 9\theta - \tan 3\theta = \dfrac{{2\sin 3\theta \cos 3\theta }}{{\cos 9\theta \cos 3\theta }}
∴tan⁡9θ−tan⁡3θ=2sin⁡3θcos⁡9θ\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⁡27θ−tan⁡9θ\tan 27\theta - \tan 9\theta
∴tan⁡27θ−tan⁡9θ=sin⁡27θcos⁡27θ−sin⁡9θcos⁡9θ\therefore \tan 27\theta - \tan 9\theta = \dfrac{{\sin 27\theta }}{{\cos 27\theta }} - \dfrac{{\sin 9\theta }}{{\cos 9\theta }}
∴tan⁡27θ−tan⁡9θ=sin⁡27θcos⁡9θ−sin⁡9θcos⁡27θcos⁡27θcos⁡9θ\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 }}
∴tan⁡27θ−tan⁡9θ=sin⁡(27θ−9θ)cos⁡27θcos⁡9θ\therefore \tan 27\theta - \tan 9\theta = \dfrac{{\sin \left( {27\theta - 9\theta } \right)}}{{\cos 27\theta \cos 9\theta }}
∴tan⁡27θ−tan⁡9θ=sin⁡18θcos⁡27θcos⁡9θ\therefore \tan 27\theta - \tan 9\theta = \dfrac{{\sin 18\theta }}{{\cos 27\theta \cos 9\theta }}
∴tan⁡27θ−tan⁡9θ=2sin⁡9θcos⁡9θcos⁡27θcos⁡9θ\therefore \tan 27\theta - \tan 9\theta = \dfrac{{2\sin 9\theta \cos 9\theta }}{{\cos 27\theta \cos 9\theta }}
∴tan⁡27θ−tan⁡9θ=2sin⁡9θcos⁡27θ\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,
∴(tan⁡27θ−tan⁡9θ)+(tan⁡9θ−tan⁡3θ)+(tan⁡3θ−tan⁡θ)=2sin⁡9θcos⁡27θ+2sin⁡3θcos⁡9θ+2sin⁡θcos⁡3θ\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 }}
∴(tan⁡27θ−tan⁡θ)=2(sin⁡9θcos⁡27θ+sin⁡3θcos⁡9θ+sin⁡θcos⁡3θ)\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)
∴(tan⁡27θ−tan⁡θ)=2(sin⁡θcos⁡3θ+sin⁡3θcos⁡9θ+sin⁡9θcos⁡27θ)\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 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 }}
∴k1=2k2\therefore {k_1} = 2{k_2}

Hence, option (A) is correct.

Note:
As sin⁡,cos⁡\sin ,\cos and tan⁡\tan functions are the most commonly used trigonometry ratio.
Some properties containing functions of sin⁡,cos⁡\sin ,\cos and tan⁡\tan are as follows:
tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta
cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta
tan⁡2θ=2tan⁡θ1+tan⁡2θ\tan 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}