 QUESTION

# If $\sin A,\cos A,\tan A$ are in G.P. then ${\cot ^6}A - {\cot ^2}A$ =?(A) -1(B) 1(C) 0(D) 2

Hint: In order to solve the problem start with the terms in the G.P. Use the property of G.P. in order to form the equation of the required form by the use of trigonometric identities such as
${\sec ^2}\theta = 1 + {\tan ^2}\theta ,\dfrac{1}{{\cos \theta }} = \sec \theta {\text{ and }}\tan \theta = \dfrac{1}{{\cot \theta }}$

Given that $\sin A,\cos A,\tan A$ are in G.P.
We have to find the value of ${\cot ^6}A - {\cot ^2}A$
As we know that if any general term $a,b,c$ are in G.P. then it can be expressed as ${b^2} = ac$
By applying this property on the given G.P. terms $\sin A,\cos A,\tan A$ we get:
$\Rightarrow {\cos ^2}A = \sin A\tan A$ --(1)
Now we will try to make ${\cot ^6}A$ from the above equation.
As we know that $\tan A = \dfrac{{\sin A}}{{\cos A}}$ ---(2)
Substituting equation (2) in equation (1) we get
$\Rightarrow {\cos ^2}A = \sin A\tan A \\ \Rightarrow {\cos ^2}A = \sin A\left( {\dfrac{{\sin A}}{{\cos A}}} \right) \\ \Rightarrow {\cos ^2}A = \dfrac{{{{\sin }^2}A}}{{\cos A}} \\ \Rightarrow \dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}} = \dfrac{1}{{\cos A}} \\$
In the above equation we have tried to bring the equation in form of $\dfrac{{\cos A}}{{\sin A}} = \cot A$
Now, we will substitute the value $\dfrac{1}{{\cos A}} = \sec A$ in the above equation, so we have:
$\Rightarrow \dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}} = \dfrac{1}{{\cos A}} = \sec A \\ \Rightarrow {\cot ^2}A = \sec A{\text{ }}\left[ {\because \dfrac{{\cos A}}{{\sin A}} = \cot A} \right] \\$
Now let us take square of both the side
$\Rightarrow {\cot ^4}A = {\sec ^2}A$
Now let us convert $\sec A{\text{ into }}\tan A$
$\Rightarrow {\cot ^4}A = {\sec ^2}A = 1 + {\tan ^2}A{\text{ }}\left[ {\because 1 + {{\tan }^2}A = {{\sec }^2}A} \right] \\ \Rightarrow {\cot ^4}A = 1 + {\left( {\tan A} \right)^2} \\ \Rightarrow {\cot ^4}A = 1 + {\left( {\dfrac{1}{{\cot A}}} \right)^2}{\text{ }}\left[ {\because \tan A = \dfrac{1}{{\cot A}}} \right] \\ \Rightarrow {\cot ^4}A = 1 + \dfrac{1}{{{{\cot }^2}A}} \\$
Now let us take LCM on the RHS and proceed to form the equation
$\Rightarrow {\cot ^4}A = \dfrac{{{{\cot }^2}A + 1}}{{{{\cot }^2}A}} \\ \Rightarrow {\cot ^4}A \times {\cot ^2}A = {\cot ^2}A + 1 \\ \Rightarrow {\cot ^{\left( {4 + 2} \right)}}A = {\cot ^2}A + 1{\text{ }}\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \\ \Rightarrow {\cot ^6}A = {\cot ^2}A + 1 \\ \Rightarrow {\cot ^6}A - {\cot ^2}A = 1 \\$
Hence, the required value of ${\cot ^6}A - {\cot ^2}A = 1$ .
So, option B is the correct option.

Note: A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In order to solve such problems, never try to find the values of the angle given like A. These types of problems can be done only by manipulation of the trigonometric identities in order to bring in the required form. Students must remember the trigonometric identities used.