Question

If \[\sin A,\cos A\] and \[\tan A\] are in G.P., then \[{\cos ^3}A + {\cos ^2}A\] is equal to
A. 1
B. 2
C. 4
D. None of these

Answer 1

Hint: For the given question we will substitute the \[\sin A,\cos A\] and \[\tan A\] in the formula of G.P. and we will substitute the value of \[\tan A\] and solve the question and will further substitute the value of \[{\sin ^2}A\] to get the correct equation.

Formula used:
The geometric mean of a and c is and b if a, b, and c are three different GP values. This might be expressed as: \[{{\rm{b}}^{\rm{2}}}{\rm{\; = ac}}\].

Complete step by step solution:
We have been given that \[\sin {\rm{A}},\cos {\rm{A}}\] and \[\tan {\rm{A}}\] are in G.P.
From the formula of G.P. which states that if a, b, c are in G.P., then \[{b^2} = ac\],
We will substitute the value of \[\sin {\rm{A}},\cos {\rm{A}}\] and \[\tan {\rm{A}}\] in the place of a, b, c.
Therefore, \[{\cos ^2}A = \sin A\tan A\] …….(1)
We will substitute the value of \[\tan A = \dfrac{{\sin A}}{{\cos A}}\] in equation (1).
Therefore,
\[{\cos ^2}A = \sin A(\dfrac{{\sin A}}{{\cos A}})\]
\[{\cos ^3}A = {\sin ^2}A\]
We will substitute the value of \[{\sin ^2}A = 1 - {\cos ^2}A\]
Therefore,
\[{\cos ^3}A = 1 - {\cos ^2}A\]
\[{\cos ^3}A + {\cos ^2}A = 1\]
Hence, option A is correct.

Additional information:
We will use the concept of geometric progression to solve the question which states that the sequence where each word is altered by another by a common ratio is known as a geometric progression or geometric sequence. When we multiply the previous term by a constant (which is not zero), we get the following term in the series. It is symbolized by:
\[a,{\rm{ }}ar,{\rm{ }}a{r^2},{\rm{ }}a{r^3},{\rm{ }}a{r^4},........\] where a is the first term and r is the common ratio.

Note: Students make mistake in substituting the value of \[\tan A\] as they might substitute the value of \[\sin A\] and make question more complex to solve and end up getting wrong answer or no answer.