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Question: Solve \[\sin \theta \sec 3\theta + \sin 3\theta \sec {3^2}\theta + \sin {3^2}\theta \sec {3^3}\theta...

Solve sin⁡θsec⁡3θ+sin⁡3θsec⁡32θ+sin⁡32θsec⁡33θ……….....nterms\sin \theta \sec 3\theta + \sin 3\theta \sec {3^2}\theta + \sin {3^2}\theta \sec {3^3}\theta ……….....n terms
A) \dfrac{1}{2}\left\\{ {\tan {3^n}\theta - \tan {3^{n - 1}}\theta } \right\\}
B) \left\\{ {\tan {3^n}\theta - \tan \theta } \right\\}
C) \dfrac{1}{2}\left\\{ {\tan {3^n}\theta - \tan \theta } \right\\}
D) None of these

Explanation

Solution

The best approach that can be followed to determine the term is to first write the question in the form of a nthn^{th} term such that we can find some trigonometric ways or identities to solve the question.

Complete step-by-step answer:
To solve the above question the first step is to write it in the form of nth term
sin⁡θsec⁡3θ+sin⁡3θsec⁡32θ+sin⁡32θsec⁡33θ………....nterms\sin \theta \sec 3\theta + \sin 3\theta \sec {3^2}\theta + \sin {3^2}\theta \sec {3^3}\theta ………....n terms
To write the above in nth term we need to write it in the form of summation because they are present in the form of sum
∑r=1nsin⁡3r−1θsec⁡3rθ\sum\limits_{r = 1}^n {\sin {3^{r - 1}}\theta \sec {3^r}\theta }
The above represent the solution in nth term now we need to find some way to simplify it and for that the best way is multiply or divide by some specific term as stated below
=∑r=1nsin⁡3r−1θsec⁡3rθ2cos⁡3r−1θ2cos⁡3r−1θ\sum\limits_{r = 1}^n {\sin {3^{r - 1}}\theta \sec {3^r}\theta } \dfrac{{2\cos {3^{r - 1}}\theta }}{{2\cos {3^{r - 1}}\theta }}
Sine, the secant function is the reciprocal of the cosine function then in such case the above can be written as
=∑r=1nsin⁡3r−1θ2cos⁡3r−1θ2cos⁡3r−1θcos⁡3rθ\sum\limits_{r = 1}^n {\sin {3^{r - 1}}\theta } \dfrac{{2\cos {3^{r - 1}}\theta }}{{2\cos {3^{r - 1}}\theta \cos {3^r}\theta }}
The above can also be written as
=12∑r=1n2cos⁡3r−1θsin⁡3r−1θcos⁡3r−1θcos⁡3rθ\dfrac{1}{2}\sum\limits_{r = 1}^n {\dfrac{{2\cos {3^{r - 1}}\theta \sin {3^{r - 1}}\theta }}{{\cos {3^{r - 1}}\theta \cos {3^r}\theta }}}
Hence, the above can be rewritten as
=12∑r=1nsin⁡(3rθ−3r−1θ)cos⁡3r−1θcos⁡3rθ\dfrac{1}{2}\sum\limits_{r = 1}^n {\dfrac{{\sin ({3^r}\theta - {3^{r - 1}}\theta )}}{{\cos {3^{r - 1}}\theta \cos {3^r}\theta }}}
And using the trigonometric relation, we get
sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b\sin (a - b) = \sin a\cos b - \cos a\sin b , in the above equation we get
=12∑r=1nsin⁡3rθcos⁡3r−1θ−cos⁡3rθsin⁡3r−1θcos⁡3r−1θcos⁡3rθ\dfrac{1}{2}\sum\limits_{r = 1}^n {\dfrac{{\sin {3^r}\theta \cos {3^{r - 1}}\theta - \cos {3^r}\theta \sin {3^{r - 1}}\theta }}{{\cos {3^{r - 1}}\theta \cos {3^r}\theta }}}
Simplifying the above, we get
=12∑r=1nsin⁡3rθcos⁡3r−1θcos⁡3r−1θcos⁡3rθ−cos⁡3rθsin⁡3r−1θcos⁡3r−1θcos⁡3rθ\dfrac{1}{2}\sum\limits_{r = 1}^n {\dfrac{{\sin {3^r}\theta \cos {3^{r - 1}}\theta }}{{\cos {3^{r - 1}}\theta \cos {3^r}\theta }}} - \dfrac{{\cos {3^r}\theta \sin {3^{r - 1}}\theta }}{{\cos {3^{r - 1}}\theta \cos {3^r}\theta }}
Hence, we know that the reciprocal of sine function and cosine function is known as tangent function so the above equation can be simplified as follows
=12∑r=1n(tan⁡3rθ−tan⁡3r−1θ)\dfrac{1}{2}\sum\limits_{r = 1}^n {\left( {\tan {3^r}\theta - \tan {3^{r - 1}}\theta } \right)}
Writing for nthn^{th} term we get
=12((tan⁡3θ−tan⁡θ)+(tan⁡32θ−tan⁡3θ)....(tan⁡3nθ−tan⁡3n−1θ)\dfrac{1}{2}\left( {\left( {\tan 3\theta - \tan \theta } \right) + \left( {\tan {3^2}\theta - \tan 3\theta } \right)....(\tan {3^n}\theta - \tan {3^{n - 1}}\theta } \right)
Hence, the above can also be written as
=12(tan⁡3nθ−tan⁡θ)\dfrac{1}{2}(\tan {3^n}\theta - \tan \theta )
Hence, the above is required solution

Note: The question involves the trigonometric function and trigonometric function contains various identities and find which identity should use in that specific part so for that we need to see what we desire for as in the above question we need the answer in tangent function or any other function as shown in the option so, we use the relation which involves sine function as well cosine function to get the result.