Answer
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Hint: We know that when three numbers \[a,b,c\] are in GP then \[{{b}^{2}}=ac\] and we have to know the basic trigonometric identity that is \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]. In the above we only have \[{{\sin }^{2}}A=\cos A\times \cot A\] then we will get \[{{\tan }^{2}}A=\cos ecA\] by dividing the above term with \[{{\cos }^{2}}A\] and proceed using trigonometric formulas.
Complete step-by-step solution:
Note: We have to know the basic trigonometric formulas that are \[\tan A=\dfrac{\sin A}{\cos A}\] and \[\cot A=\dfrac{\cos A}{\sin A}\] and the product of \[\sin A\] and \[\cos ecA\] is 1 because they are reciprocal to each other. In general the given numbers are said to be in GP if the ratio of second term to first term and ratio of third term to the second term is the same which is called the common ratio and denoted by “r”.
Complete step-by-step solution:
Given \[\cos A,\sin A,\cot A\] are in GP
We know that if three numbers \[a,b,c\] are in GP then \[{{b}^{2}}=ac\]
So by applying the formulae we will get
\[{{\sin }^{2}}A=\cos A\times \cot A\]. . . . . . . . . . . . . . (1)
Dividing with \[{{\cos }^{2}}A\] on both the left hand side and right hand side we will get,
\[\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}=\dfrac{\cos A\times \cot A}{{{\cos }^{2}}A}\]
\[{{\tan }^{2}}A=\cos ecA\]. . . . . . . . . . . . . . . . . . .(2)
Now multiply with \[\tan A\] on both left hand side and right hand side then we will get
\[{{\tan }^{2}}A\times \tan A=\cos ecA\times \tan A\]
We know that the \[\tan A\] is the ratio of \[\sin A\] to the \[\cos A\]. By applying that formula we will get
\[{{\tan }^{3}}A=\cos ecA\times \dfrac{\sin A}{\cos A}\]
\[{{\tan }^{3}}A=\sec A\]. . . . . . . . . . . . . . . . . . . . . . (3)
Now squaring on both left hand side and right hand side we will get
\[{{\left( {{\tan }^{3}}A \right)}^{2}}={{\sec }^{2}}A\]
\[{{\tan }^{6}}A={{\sec }^{2}}A\]. . . . . . . . . . . . . . . . . . . . (4)
We want to know the value of \[{{\tan }^{6}}A-{{\tan }^{2}}A\], So subtract \[{{\tan }^{2}}A\], from both the sides.
\[{{\tan }^{6}}A-{{\tan }^{2}}A={{\sec }^{2}}A-{{\tan }^{2}}A\]
We know the trigonometric identity that \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]
So, \[{{\tan }^{6}}A-{{\tan }^{2}}A=1\]
So, the correct option for above question is option (C).
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