Question

If $\sin\theta , \cos \theta $ and $ \tan \theta$ are in G.P, then $\cot^6 \theta - \cot^2 \theta$ is

• A. 1
• B. 1/2
• C. 2
• D. 3

Answer 1

Answer: (A)

$\cos^{2}\theta =\sin\theta \tan\theta \Rightarrow \cos^{3}\theta=\sin^{2}\theta $
$\Rightarrow \cot^{2 }\theta =\sec\theta \cot^{6} \theta-\cot^{2} \theta =\cot^{2} \theta\left(\cot^{4} \theta -1\right)$
$= \cot^{2} \theta \cdot cosec^{2} \theta\left(\cot^{2} \theta -1\right)$
$=\sec\theta \cdot \frac{1}{\sin^{2} \theta}\left(-\sec\theta-1\right)$
$= \frac{\sec \theta}{1-\cos^{2} \theta}\cdot \left(\frac{1 -\cos\theta }{\cos \theta}\right) $
$= \frac{\sec^{2} \theta}{1+\cos \theta}$
$\cos^{2} \theta =\sin\theta \tan \theta \Rightarrow \cos^{3}\theta = \sin^{2}\theta$
$ \cot^{6} \theta - \cot^{2} \theta =\frac{\left(\cos^{3}\theta\right)^{2}}{\sin^{6}\theta}-\frac{\cos^{2}\theta}{\sin^{2}\theta}$
$= \frac{\sin^{4} \theta}{\sin^{6} \theta} - \frac{\cos^{2} \theta}{\cos^{3} \theta}$
$= \frac{1 }{\sin^{2} \theta} - \frac{1}{\cos\:\theta}$
$= \frac{\cos\: \theta - \sin^2 \theta }{\sin^{2} \theta \: \cos \theta}$
$= \frac{\cos\: \theta - \cos^3\theta }{\sin^{2} \theta \: \cos \theta}$
$= \frac{\cos\: \theta (1- \cos^2 \theta )}{\sin^{2} \theta \: \cos \theta} = 1$