Question

If $\sin \theta ,\cos \theta$ and $\tan \theta$ are in GP then ${\cot }^6\theta -{\cot }^2\theta =$

• A. $4$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

Since $\sin \theta ,\cos \theta ,and\tan \theta$ are in GP
${\cos }^2\theta =\sin \theta \tan \theta$
$=\sin \theta \frac{\sin \theta }{\cos \theta }$
or ${\cos }^3\theta ={\sin }^2\theta$
or$\frac{{\cos }^3\theta }{{\sin }^2\theta }=1$
or$\frac{{\cos }^3\theta }{{\sin }^3\theta }=\frac{1}{\sin \theta }$
or ${\cot }^3\theta =\text{cosec}\theta$
$\therefore {\cot }^6\theta -{\cot }^2\theta ={\text{cosec}}^2\theta -({\text{cosec}}^2\theta -1)=1$