Question

If $\sin A,\cos A,\tan A$ are $3$ consecutive terms of a $G.P$ then what can you say about $({\cot }^{6}A-{\cot }^{2}A)$?

Answer 1

If $\sin A,\cos A$ and $\tan A$ are in $G.P$.

$\cos A/\sin A=\tan A/\cos A$.

${\cos }^{2}A=\tan A.\sin A$.

${\cos }^{3}A={\sin }^{2}A$.

$\cos A/sinA=\tan A/\cos A$

$\cot A=1/\cot A.1/\cos A$

${\cot }^{2}A=1/\cos A=\sec A$

${\cot }^{6}A={\sec }^{3}A$

$={\cot }^{6}A-{\cot }^{2}A$

$={\sec }^{3}A-\sec A$

$=\sec A({\sec }^{2}A-1)$

$=\sec A.{\tan }^{2}A$

$=\frac{1}{\cos A}.\frac{{\sin }^{2}A}{{\cos }^{2}A}$.

$=\frac{{\sin }^{2}A}{{\cos }^{3}A}$

as ${\sin }^{2}A={\cos }^{3}A$.

${\cot }^{6}A-{\cot }^{2}A=1$