Question

If ${{a}^{2}}$, ${{b}^{2}}$ and ${{c}^{2}}$are in A.P., then which of the following is also an A.P.?
A.$\sin A,\sin B,\sin C$
B.$\tan A,\tan B,\tan C$
C.$\cot A,\cot B,\cot C$
D.None of these

Answer 1

Hint : Assume ${{a}_{1}},{{a}_{2}},{{a}_{3}},.....,{{a}_{n}}$ are in A.P. if ${{a}_{2}}={{a}_{1}}+d$, ${{a}_{3}}={{a}_{2}}+d$, and so on. Here constant $d$ is the common difference of an A.P. In other words, a sequence or a series ${{a}_{1}},{{a}_{2}},{{a}_{3}},.....,{{a}_{n}}$is called arithmetic progression if the difference of a term and the preceding term is always constant. Let $a$ be the first term and $d$ be the common difference of an A.P. Then. Its ${{n}^{th}}$ term or general term is given by
General term of an A.P. = First term + (Term number-1) × (Common Difference)
Further equating we get the formula as
${{a}_{n}}=a+(n-1)d$
If three numbers $a,b,c$ in order are in A.P. Then, $2b=a+c$
If $a,b,c$ are in A.P., then $b$ is called the arithmetic mean (AM) between $a$ and $c$, that is $b=\dfrac{a+c}{2}$
The sum ${{S}_{n}}$ of$n$ terms of an A.P. with first term and common difference is
{{S}_{n}}=\dfrac{n}{2}\left\\{ 2a+(n-1)d \right\\}

Complete step-by-step answer :
According to the question,
${{a}^{2}}$, ${{b}^{2}}$ and ${{c}^{2}}$are in A.P.
Therefore, the common difference between them will be equal. So we can equate them as
${{b}^{2}}-{{a}^{2}}={{c}^{2}}-{{b}^{2}}$
Here we need to apply the sine law. According to the law,
$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$
From the above equation we get
$a=k\sin A$,
$b=k\sin B$
And $c=k\sin C$
Substituting these values we get
${{k}^{2}}{{\sin }^{2}}B-{{k}^{2}}{{\sin }^{2}}A={{k}^{2}}{{\sin }^{2}}C-{{k}^{2}}{{\sin }^{2}}B$
Taking ${{k}^{2}}$common on both sides we get
${{k}^{2}}({{\sin }^{2}}B-{{\sin }^{2}}A)={{k}^{2}}({{\sin }^{2}}C-{{\sin }^{2}}B)$
Cancelling ${{k}^{2}}$ from both sides we get
${{\sin }^{2}}B-{{\sin }^{2}}A={{\sin }^{2}}C-{{\sin }^{2}}B$
As we know the trigonometry formula
${{\sin }^{2}}A-{{\sin }^{2}}B=\sin (A+B)\sin (A-B)$
Applying the above formula we get
$\sin (B+A)\sin (B-A)=\sin (C+B)\sin (C-B)$
As we know that $A,B,C$ are the angles of the right triangle and their sum is $180$. We can write it as $A+B+C=\pi$
Hence we can further rewrite the equation as
$\sin (\pi -C)\sin (B-A)=\sin (\pi -A)\sin (C-B)$
Further solving we get
$\sin C\sin (B-A)=\sin A\sin (C-B)$
Rearranging the equation we get
$\dfrac{\sin (B-A)}{\sin A}=\dfrac{\sin (C-B)}{\sin C}$
Dividing both sides by $\sin B$ we get
$\dfrac{\sin (B-A)}{\sin A\sin B}=\dfrac{\sin (C-B)}{sinB\sin C}$
Using trigonometric formula we get
$\dfrac{\sin B\cos A-\cos B\sin A}{\sin A\sin B}=\dfrac{\sin C\cos B-\cos C\sin B}{\sin B\sin C}$
Further simplifying we get
$\cot A-\cot B=\cot B-\cot C$
Therefore, $\cot A,\cot B,\cot C$ are in A.P.
Thus option $\left( C \right)$is the correct answer.
So, the correct answer is “Option C”.

Note : A sequence or a series is an arrangement of numbers in a definite border according to some pattern. These types of questions require the knowledge of Arithmetic progression and trigonometric concepts.