Question

If $\sin \theta$, $\cos \theta$ and $\tan \theta$ are in G.P., then prove that ${\cot }^6\theta -{\cot }^2\theta =1$.

Answer 1

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

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

$⟹{\cos }^3\theta ={\sin }^2\theta$

$⟹{\cot }^3\theta =\frac{1}{\sin \theta }=\csc \theta$

Squaring on both sides, we get

${\cot }^6\theta ={\csc }^2\theta$

$⟹{\cot }^6\theta =1+{\cot }^2\theta$

Hence proved.

$\therefore {\cot }^6\theta -{\cot }^2\theta =1$