Question

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

• A. $1$
• B. $\frac{1}{2}$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$1$

Given that $\sin \theta ,\cos \theta$ and $\tan \theta$ are in G.P

So, we have

${\cos }^2\theta =\tan \theta \sin \theta$

$\Rightarrow {\cos }^3\theta ={\sin }^2\theta$ .... $\left(i\right)$

Now,

${\cot }^6\theta -{\cot }^2\theta ={\cot }^2\theta ({\cot }^4\theta -1)=\frac{{\cos }^2\theta }{{\sin }^2\theta }(\frac{{\cos }^4\theta }{{\sin }^4\theta }-1)$

$=\frac{{\cos }^2\theta }{{\cos }^3\theta }(\frac{{\cos }^4\theta }{{\cos }^6\theta }-1)$ ...... From $\left(i\right)$

$=\frac{1}{\cos \theta }\frac{{\sin }^2\theta }{{\cos }^2\theta }=\frac{{\cos }^3\theta }{{\cos }^3\theta }=1$