Question

If $\sin A,\cos A$ and $\tan A$ are in Geometric progression, then ${\cot }^{6}A-{\cot }^{2}A$ is

• A. $2$
• B. $4$
• C. $3$
• D. $1$

Answer 1

The correct option is D $1$

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

To find out- ${\cot }^{6}A-{\cot }^{2}A=?$

Solution-

$\frac{\sin A}{\cos A}=\frac{\cos A}{\tan A}$

$\Rightarrow \tan A=\cot A\cos A=\cot A\times \frac{1}{\sec A}$

$\Rightarrow \frac{1}{\cot A}=\cot A\times \frac{1}{\sec A}$

$\Rightarrow {\cot }^{2}A=\sec A$

$\Rightarrow {\cot }^{4}A={\sec }^{2}A$

$\Rightarrow {\cot }^{4}A=1+{\tan }^{2}A$

$\Rightarrow {\cot }^{4}A=1+\frac{1}{{\cot }^{2}A}$

$\Rightarrow {\cot }^{6}A={\cot }^{2}A+1$

$\Rightarrow {\cot }^{6}A-{\cot }^{4}A=1$