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.