Question

If cosA, sinA, cotA are in GP then ${\tan }^{6}A-{\tan }^{2}A=$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

Given, $\cos A,\sin A,\cot A$ are in G.P. then we've,

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

or, ${\tan }^{2}A=\cos ecA$

Now multiplying both sides of the above equation by $\tan A$ then we get,

or, ${\tan }^{3}A=\sec A$.....(1).

Now,

${\tan }^{6}A-{\tan }^{2}A$

$=(\sec A{)}^2-{\tan }^{2}A$ [ Using (1)]

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

$=1$.