Question

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

• A. \(\frac1{81}\)
• B. \(\frac{1}{9}\)
• C. \(\frac{1}{27}\)
• D. \(\frac{1}{243}\)
• E. None of the above

Answer 1

The Correct Option is C

\(1+\frac{\tanθ}{\cotθ}+\bigg(\frac{\tanθ}{\cotθ}\bigg)^2+\bigg(\frac{\tanθ}{\cotθ}\bigg)^3+...∞ = \frac32\)

\(\Rightarrow 1+\frac{\tan\theta}{\frac{1}{\tan\theta}}+\bigg(\frac{\tan\theta}{\frac{1}{\tan\theta}}\bigg)^2+\bigg(\frac{\tan\theta}{\frac{1}{\tan\theta}}\bigg)^3+...\infin=\frac{3}{2}\)

\(\Rightarrow1+\tan^2\theta+\tan^4\theta+\tan^6\theta+...\infin=\frac{3}{2}\)

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, \(\frac{1}{1-(\tan\theta)^2}=\frac{3}{2}\)

\(2=3-3(\tan\theta)^2\)

\(\tan^2\theta=\frac{1}{3}\)

Thus, \(\frac{\tan^2\theta-\tan^4\theta}{\cot^4\theta-\cot^2\theta}\)

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

\(=\frac{\frac{1}{3}-\frac{1}{9}}{9-3}\)

\(=\frac{1}{27}\)

Hence, option C is the correct answer.