Question

If $\sin A,\cos A$ & an $A$ are in $GP$ then ${\cot }^{6}A-{\cot }^{2}A$ is equal to

• A. $-1$
• B. $0$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $1$

$\sin A,\cos A\&\tan A$ are in

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

$\Rightarrow {\cos }^{2}A=\sin A\cdot \frac{\sin A}{\cos A}$

$\Rightarrow \frac{{\cos }^{2}A}{{\sin }^{2}A}=\frac{1}{\cos A}⟶\left(i\right)$

$\Rightarrow {\cot }^{2}A=\sec A-\left(i\right)$

Cubing both sides,

$\Rightarrow {\cot }^{6}A={\sec }^{3}A-$ (iii)

Now, ${\cot }^{6}A-{\cot }^{2}A$

$={\sec }^{3}A-\sec A$ ( from (ii) $\&($ iii $)$

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

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

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

$=\frac{{\sin }^{2}A}{{\cos }^{3}A}=1($ from $\left(i\right))$