Question

If $\sin A,\cos A$ and $\tan A$ are in G.P. 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 $1$

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

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

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

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

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

Squaring on both sides, we get

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

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

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

Now, ${\cot }^{6}A-{\cot }^{2}A={\cot }^{2}A({\cot }^{4}A-1)$

$={\cot }^{2}A\times {\tan }^{2}A$

$=1$