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)

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