Solveeit Logo
Login

Question

Question: If \(\cot A + \cot B + \cot C = \sqrt 3 \) then the triangle ABC is- A.Equilateral B.Right angle...

If cot⁡A+cot⁡B+cot⁡C=3\cot A + \cot B + \cot C = \sqrt 3 then the triangle ABC is-
A.Equilateral
B.Right angled
C.Isosceles
D.None of these

Explanation

Solution

First, we will find the square of the given equation. Then we know that the sum of the angles of the triangle is 180∘{180^\circ } so we will use this theorem and apply cot in the equation formed. Then use the formula-cot⁡(A+B)=cot⁡Acot⁡B−1cot⁡A+cot⁡B\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}} and form an equation. Put the value of this equation in the equation obtained before. Then solve to get the identity (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} -2ab and then equate the terms to zero and solve again to get the answer.

Complete step-by-step answer:
Given, cot⁡A+cot⁡B+cot⁡C=3\cot A + \cot B + \cot C = \sqrt 3 --- (i)
Now we know that the sum of the angles of the triangle is 180∘{180^\circ }
In △ABC\vartriangle ABC, the angles are A, B and C then on applying the above theorem, we get-
⇒A+B+C=180∘\Rightarrow A + B + C = {180^\circ }
So we can write-
⇒\Rightarrow A+B=180∘−CA + B = {180^ \circ } - C
On applying cot both side, we get-
⇒cot⁡(A+B)=cot⁡(180∘−C)\Rightarrow \cot \left( {A + B} \right) = \cot \left( {{{180}^ \circ } - C} \right)--- (ii)
Now, we know that the formula of cot⁡(A+B)=cot⁡Acot⁡B−1cot⁡A+cot⁡B\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}
On applying this in eq. (ii), we get-
⇒cot⁡Acot⁡B−1cot⁡A+cot⁡B=cot⁡(180∘−C)\Rightarrow \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}} = \cot \left( {{{180}^\circ } - C} \right)
Now we know thatcot⁡(180∘−θ)=−cot⁡θ\cot \left( {{{180}^ \circ } - \theta } \right) = - \cot \theta , so we get-
⇒cot⁡Acot⁡B−1cot⁡A+cot⁡B=−cot⁡C\Rightarrow \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}} = - \cot C
On cross multiplication, we get-
⇒cot⁡A.cot⁡B−1=−cot⁡Acot⁡C−cot⁡Ccot⁡B\Rightarrow \cot A.\cot B - 1 = - \cot A\cot C - \cot C\cot B
On rearranging, we get-
⇒cot⁡A.cot⁡B+cot⁡Ccot⁡B+cot⁡Acot⁡C=1\Rightarrow \cot A.\cot B + \cot C\cot B + \cot A\cot C = 1--- (iii)
Now on squaring, eq. (i), we get-
⇒(cot⁡A+cot⁡B+cot⁡C)2=(3)2\Rightarrow {\left( {\cot A + \cot B + \cot C} \right)^2} = {\left( {\sqrt 3 } \right)^2}
On using formula (a+b+c)2=a2+b2+c2+2(ab+bc+ca){\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) , we get-
⇒cot⁡2A+cot⁡2B+cot⁡2C+2(cot⁡Acot⁡B+cot⁡Bcot⁡C+cot⁡Ccot⁡A)=3\Rightarrow {\cot ^2}A + {\cot ^2}B + {\cot ^2}C + 2\left( {\cot A\cot B + \cot B\cot C + \cot C\cot A} \right) = 3--- (iv)
Now on substituting value from eq. (iii) in eq. (iv), we get-
⇒cot⁡2A+cot⁡2B+cot⁡2C+2×1=3\Rightarrow {\cot ^2}A + {\cot ^2}B + {\cot ^2}C + 2 \times 1 = 3
⇒cot⁡2A+cot⁡2B+cot⁡2C+2=3\Rightarrow {\cot ^2}A + {\cot ^2}B + {\cot ^2}C + 2 = 3
On transferring the constant on other side, we get-
⇒cot⁡2A+cot⁡2B+cot⁡2C=3−2\Rightarrow {\cot ^2}A + {\cot ^2}B + {\cot ^2}C = 3 - 2
On simplifying, we get-
⇒cot⁡2A+cot⁡2B+cot⁡2C=1\Rightarrow {\cot ^2}A + {\cot ^2}B + {\cot ^2}C = 1
Now, on substituting the value from eq. (iii) and multiplying the eq. by 2, we get-
⇒2cot⁡2A+2cot⁡2B+2cot⁡2C=2(cot⁡Acot⁡B+cot⁡Acot⁡C+cot⁡Bcot⁡C)\Rightarrow 2{\cot ^2}A + 2{\cot ^2}B + 2{\cot ^2}C = 2\left( {\cot A\cot B + \cot A\cot C + \cot B\cot C} \right)
On rearranging, we get-
⇒2cot⁡2A+2cot⁡2B+2cot⁡2C−2cot⁡Acot⁡B−2cot⁡Acot⁡C−2cot⁡Bcot⁡C=0\Rightarrow 2{\cot ^2}A + 2{\cot ^2}B + 2{\cot ^2}C - 2\cot A\cot B - 2\cot A\cot C - 2\cot B\cot C = 0
Now we can write the eq. as-
⇒cot⁡2A+cot⁡2A+cot⁡2B+cot⁡2B+cot⁡2C+cot⁡2C−2cot⁡Acot⁡B−2cot⁡Acot⁡C−2cot⁡Bcot⁡C=0\Rightarrow {\cot ^2}A + {\cot ^2}A + {\cot ^2}B + {\cot ^2}B + {\cot ^2}C + {\cot ^2}C - 2\cot A\cot B - 2\cot A\cot C - 2\cot B\cot C = 0On again rearranging, we get-
⇒(cot⁡2A+cot⁡2B−2cot⁡Acot⁡B)+(cot⁡2A+cot⁡2C−2cot⁡Acot⁡C)+(cot⁡2B+cot⁡2C−2cot⁡Bcot⁡C)=0\Rightarrow \left( {{{\cot }^2}A + {{\cot }^2}B - 2\cot A\cot B} \right) + \left( {{{\cot }^2}A + {{\cot }^2}C - 2\cot A\cot C} \right) + \left( {{{\cot }^2}B + {{\cot }^2}C - 2\cot B\cot C} \right) = 0Now we know that (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab
On applying this formula, we get-
⇒(cot⁡A−cot⁡B)2+(cot⁡B−cot⁡C)2+(cot⁡C−cot⁡A)2=0\Rightarrow {\left( {\cot A - \cot B} \right)^2} + {\left( {\cot B - \cot C} \right)^2} + {\left( {\cot C - \cot A} \right)^2} = 0--- (v)
Then we get-
⇒(cot⁡A−cot⁡B)2=0 and (cot⁡B−cot⁡C)2=0 and (cot⁡C−cot⁡A)2=0\Rightarrow {\left( {\cot A - \cot B} \right)^2} = 0{\text{ and }}{\left( {\cot B - \cot C} \right)^2} = 0{\text{ and }}{\left( {\cot C - \cot A} \right)^2} = 0
Then we get-
⇒cot⁡A−cot⁡B=0 andcot⁡B−cot⁡C=0 and cot⁡C−cot⁡A=0\Rightarrow \cot A - \cot B = 0{\text{ and}}\cot B - \cot C = 0{\text{ and }}\cot C - \cot A = 0
On simplifying, we get-
⇒cot⁡A=cot⁡B andcot⁡B=cot⁡C and cot⁡C=cot⁡A\Rightarrow \cot A = \cot B{\text{ and}}\cot B = \cot C{\text{ and }}\cot C = \cot A
Then, we get-
⇒A=B andB=C and C=A\Rightarrow A = B{\text{ and}}B = C{\text{ and }}C = A
Hence we can write,
⇒A=B=C\Rightarrow A = B = C
The triangle in which all the angles are equal is an equilateral triangle.
The correct answer is option A.

Note: Here remember that the terms are in addition and not multiplication so student may go wrong if he/she writes-
⇒(cot⁡A−cot⁡B)2=0 or (cot⁡B−cot⁡C)2=0 or (cot⁡C−cot⁡A)2=0\Rightarrow {\left( {\cot A - \cot B} \right)^2} = 0{\text{ or }}{\left( {\cot B - \cot C} \right)^2} = 0{\text{ or }}{\left( {\cot C - \cot A} \right)^2} = 0 (In multiplication on term as to be zero to make the equation equal to zero)
Because then only two angles will be equal which will give us an isosceles triangle. So since there is an additional sign between the terms so all the terms will be equal to zero.