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Question: In any \[\Delta ABC\], if \[{a^2},{b^2},{c^2}\] are in A.P., then prove that \[\cot A,\cot B,\cot C\...

In any ΔABC\Delta ABC, if a2,b2,c2{a^2},{b^2},{c^2} are in A.P., then prove that cot⁡A,cot⁡B,cot⁡C\cot A,\cot B,\cot C are in A.P.

Explanation

Solution

Here, we have to prove that the cotangents of Angles are in A.P. First from the given we have to use the concept of arithmetic progression and by using the trigonometric theorem, we have to prove that the cotangents are also in A.P.. Arithmetic progression is a sequence of numbers in which the difference of any two adjacent terms is constant.

Formula Used:
We will use the following formula

  1. d=a(n)−a(n−1)d = a(n) - a(n - 1) where a(n),a(n−1)a(n),a(n - 1) are the last term and previous term respectively.
  2. Sine rule: asin⁡A=bsin⁡B=csin⁡C=k\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k
  3. The difference of square of numbers is given by the algebraic identity: a2−b2=(ab)(a−b){a^2} - {b^2} = \left( {a b} \right)\left( {a - b} \right)
  4. Trigonometric Formulae: sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B
  5. Trigonometric Identity: sin⁡(π−C)=sin⁡C;sin⁡(π−A)=sin⁡A;\sin (\pi - C) = \sin C;\sin (\pi - A) = \sin A;
  6. Trigonometric Ratio: cos⁡Asin⁡A=cot⁡A\dfrac{{\cos A}}{{\sin A}} = \cot A

Complete step by step solution:
We are given that any \Delta ABC$$$${a^2}, {b^2}, {c^2} are in A.P.
Since the terms are in A.P., the common difference is constant.
d=a(n)−a(n−1)d = a(n) - a(n - 1) where a(n),a(n−1)a(n),a(n - 1) are the last term and previous term respectively.
By using the common difference formula, we will get
⇒b2−a2=c2−b2\Rightarrow {b^2} - {a^2} = {c^2} - {b^2}
Since the given terms a,b,ca,b,c are the sides of the triangle, we will be using the sine rule
Sine rule: asin⁡A=bsin⁡B=csin⁡C=k\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k, by using the sine rule, we get
a=ksin⁡A;b=ksin⁡B;c=ksin⁡Ca = k\sin A;b = k\sin B;c = k\sin C
Substituting the values of a,b,ca,b,cin the equation, we get
⇒k2sin2B−k2sin2A=k2sin2C−k2sin2B  \Rightarrow {k^2}si{n^2}B - {k^2}si{n^2}A = {k^2}si{n^2}C - {k^2}si{n^2}B\;
Taking the common factor on both the sides, we get
⇒k2(sin2B−sin2A)=k2(sin2C−sin2B)  \Rightarrow {k^2}\left( {si{n^2}B - si{n^2}A} \right) = {k^2}\left( {si{n^2}C - si{n^2}B} \right)\;
Cancelling on both the sides, we get
⇒(sin2B−sin2A)=(sin2C−sin2B)  \Rightarrow \left( {si{n^2}B - si{n^2}A} \right) = \left( {si{n^2}C - si{n^2}B} \right)\;
The difference of square of numbers is given by the algebraic identity: a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
Now, by using the algebraic identity, we get
⇒sin⁡(B+A)sin⁡(B−A)=sin⁡(C+B)  sin⁡(C−B)\Rightarrow \sin \left( {B + A} \right)\sin \left( {B - A} \right) = \sin \left( {C + B} \right)\;\sin \left( {C - B} \right)
Since the given is a triangle, we have A+B+C=πA + B + C = \pi
So, we have A+B=π−C;B+C=π−A;A + B = \pi - C;B + C = \pi - A;
⇒sin⁡(π−C)sin⁡(B−A)=sin⁡(π−A)sin⁡(C−B)\Rightarrow \sin (\pi - C)\sin (B - A) = \sin (\pi - A)\sin (C - B)
Now, by using the Trigonometric Identity: sin⁡(π−C)=sin⁡C;sin⁡(π−A)=sin⁡A;\sin (\pi - C) = \sin C;\sin (\pi - A) = \sin A;, we get
⇒sin⁡Csin⁡(B−A)=sin⁡Asin⁡(C−B)\Rightarrow \sin C\sin (B - A) = \sin A\sin (C - B)
Rewriting the equation, we get
⇒sin⁡(B−A)sin⁡A=sin⁡(C−B)sin⁡C\Rightarrow \dfrac{{\sin (B - A)}}{{\sin A}} = \dfrac{{\sin (C - B)}}{{\sin C}}
Dividing by sin⁡B\sin Bon both the sides, we get
⇒sin⁡(B−A)sin⁡Asin⁡B=sin⁡(C−B)sin⁡Bsin⁡C\Rightarrow \dfrac{{\sin (B - A)}}{{\sin A\sin B}} = \dfrac{{\sin (C - B)}}{{\sin B\sin C}}
Now, by using the trigonometric formulae: sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B, we get
⇒sin⁡Bcos⁡A−cos⁡Bsin⁡Asin⁡Asin⁡B=sin⁡Ccos⁡B−cos⁡Bsin⁡Csin⁡Bsin⁡C\Rightarrow \dfrac{{\sin B\cos A - \cos B\sin A}}{{\sin A\sin B}} = \dfrac{{\sin C\cos B - \cos B\sin C}}{{\sin B\sin C}}
Dividing the terms on both the sides separately, we get
⇒sin⁡Bcos⁡Asin⁡Asin⁡B−cos⁡Bsin⁡Asin⁡Asin⁡B=sin⁡Ccos⁡Bsin⁡Bsin⁡C−cos⁡Bsin⁡Csin⁡Bsin⁡C\Rightarrow \dfrac{{\sin B\cos A}}{{\sin A\sin B}} - \dfrac{{\cos B\sin A}}{{\sin A\sin B}} = \dfrac{{\sin C\cos B}}{{\sin B\sin C}} - \dfrac{{\cos B\sin C}}{{\sin B\sin C}}
⇒cos⁡Asin⁡A−cos⁡Bsin⁡B=cos⁡Bsin⁡B−cos⁡Bsin⁡B\Rightarrow \dfrac{{\cos A}}{{\sin A}} - \dfrac{{\cos B}}{{\sin B}} = \dfrac{{\cos B}}{{\sin B}} - \dfrac{{\cos B}}{{\sin B}}
Now, by using the trigonometric ratio cos⁡Asin⁡A=cot⁡A\dfrac{{\cos A}}{{\sin A}} = \cot A, we get
⇒cot⁡A−cot⁡B=cot⁡B−cot⁡C\Rightarrow \cot A - \cot B = \cot B - \cot C
This is of the form d=a(n)−a(n−1)d = a(n) - a(n - 1), so this forms an A.P.

Therefore, cot⁡A,cot⁡B,cot⁡C\cot A,\cot B,\cot C are in A.P.

Note:
We should know the law of sine and its limitations. This formula can be used only when a ratio is made, meaning you must have an angle and two sides given, and one of the two sides must be opposite the given angle. Since the given is a triangle, we are using the law of sine. The Sine Rule can be used in any triangle where a side and its opposite angle are known. You will only ever need two parts of the Sine Rule formula, not all three.