First slide

Trigonometric Identities

Question

If $\sin A, \cos A$ and $\tan A$ are in G.P., then
$\cot^{6} A - \cot^{2} A =$

Moderate

A

-1
B

0
C

1
D

none of these

Solution

$\begin{array}{l} \cos^{2} A = \sin A \tan A = \sin^{2} A \cdot \sec A \\ \Rightarrow \cot^{2} A = \sec A \\ \begin{aligned} ∴ \cot^{6} A - \cot^{2} A & = \cot^{2} A (\cot^{4} A - 1) \\ = \cot^{2} A (\sec^{2} A - 1) \\ = \cot^{2} A \times \tan^{2} A = 1 \end{aligned} \end{array}$

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