Question

If $\sin a,\cos a,\tan a$ are in G.P then ${\cot }^{6}a-{\cot }^{2}a$ is equal to

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

The correct option is A $1$

Given,

$\sin a,\cos a,\tan a$ are in GP.

$\Rightarrow \frac{\cos a}{\sin a}=\frac{\tan a}{\cos a}$ ............ (i)

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

and from (i), $\cot a=\frac{1}{\cot a}\times \frac{1}{\cos a}$

$\Rightarrow {\cot }^{2}a=\sec a$........(1)

$\therefore \Rightarrow {\cot }^{6}a={\sec }^{3}a$.......(2)

(2)-(1) gives,

${\cot }^{6}a-{\cot }^{2}a={\sec }^{3}a-\sec a$

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

$=\sec a{\tan }^{2}a$

$=\frac{1}{\cos a}\times \frac{\sin a}{\cos a}$

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

but from (A), we have,

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

$\therefore {\cot }^{6}a-{\cot }^{2}a=\frac{{\sin }^{2}a}{{\cos }^{3}a}=1$