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Quantitative Ability and Data Interpretation Question on Trigonometry

If 1+tan⁡θcot⁡θ+(tan⁡θcot⁡θ)2+(tan⁡θcot⁡θ)3+...∞=321+\frac{\tanθ}{\cotθ}+\bigg(\frac{\tanθ}{\cotθ}\bigg)^2+\bigg(\frac{\tanθ}{\cotθ}\bigg)^3+...∞ = \frac32, then find the value of tan⁡2θ−tan⁡4θcot⁡4θ−cot⁡2θ\frac{\tan^2θ-\tan^4θ}{\cot^4θ-\cot^2θ}.

A

181\frac1{81}

B

19\frac{1}{9}

C

127\frac{1}{27}

D

1243\frac{1}{243}

E

None of the above

Answer

127\frac{1}{27}

Explanation

Solution

1+tan⁡θcot⁡θ+(tan⁡θcot⁡θ)2+(tan⁡θcot⁡θ)3+...∞=321+\frac{\tanθ}{\cotθ}+\bigg(\frac{\tanθ}{\cotθ}\bigg)^2+\bigg(\frac{\tanθ}{\cotθ}\bigg)^3+...∞ = \frac32

⇒1+tan⁡θ1tan⁡θ+(tan⁡θ1tan⁡θ)2+(tan⁡θ1tan⁡θ)3+...∞=32\Rightarrow 1+\frac{\tan\theta}{\frac{1}{\tan\theta}}+\bigg(\frac{\tan\theta}{\frac{1}{\tan\theta}}\bigg)^2+\bigg(\frac{\tan\theta}{\frac{1}{\tan\theta}}\bigg)^3+...\infin=\frac{3}{2}

⇒1+tan⁡2θ+tan⁡4θ+tan⁡6θ+...∞=32\Rightarrow1+\tan^2\theta+\tan^4\theta+\tan^6\theta+...\infin=\frac{3}{2}

We know that the square of any number is positive. So, from the above equation we can conclude that it is a descending GP series.

So, 11−(tan⁡θ)2=32\frac{1}{1-(\tan\theta)^2}=\frac{3}{2}

2=3−3(tan⁡θ)22=3-3(\tan\theta)^2

tan⁡2θ=13\tan^2\theta=\frac{1}{3}

Thus, tan⁡2θ−tan⁡4θcot⁡4θ−cot⁡2θ\frac{\tan^2\theta-\tan^4\theta}{\cot^4\theta-\cot^2\theta}

= 13−(13)2(3)2−3\frac{\frac{1}{3}-\bigg(\frac{1}{3}\bigg)^2}{(3)^2-3}

=13−199−3=\frac{\frac{1}{3}-\frac{1}{9}}{9-3}

=127=\frac{1}{27}

Hence, option C is the correct answer.