If sin A, cos A and tan A are in geometric progression, then $c o t^{6} A - c o t^{2} A$ is equal to

-1

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none of these

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Solution

The correct option is B 1
Since sin A, cos A and tan A are in G.P., we have

$c o s^{2} A = s i n A t a n A \Rightarrow c o s^{3} A = s i n^{2} A$

$\Rightarrow \frac{c o s^{2} A}{s i n^{2} A} = \frac{1}{c o s A} \Rightarrow c o t^{2} A = s e c A$

$\Rightarrow c o t^{4} A = s e c^{2} A = 1 + t a n^{2} A$

$\Rightarrow c o t^{4} A = 1 + \frac{1}{c o t^{2} A} \Rightarrow c o t^{6} A - c o t^{2} A = 1$

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