Question

In $\Delta \,ABC$, ${a^2} + {c^2} = 2002{b^2}$, then $\dfrac{{\cot A + \cot C}}{{\cot B}}$ equals to
A. $\dfrac{1}{{2001}}$
B. $\dfrac{2}{{2001}}$
C. $\dfrac{3}{{2001}}$
D. $\dfrac{4}{{2001}}$

Answer 1

Hint : Here in this question, we have to find the value of the given trigonometric ratio $\dfrac{{\cot A + \cot C}}{{\cot B}}$. To solve this, by the definition of trigonometric ratios we have to rewrite a $\cot \theta$ as $\dfrac{{\cos \theta }}{{\sin \theta }}$ and further substitute $\cos \theta$ and $\sin \theta$ by the law of cosine and sine ratio and on simplification we get the required solution.

Complete step by step solution:

Consider, triangle $\Delta \,ABC$, a, b, and c are sides of the triangle whereas A, B, and C are angles of triangle $\Delta \,ABC$.
In a triangle, side “a” divided by the sine of angle A is equal to the side “b” divided by the sine of angle B is equal to the side “c” divided by the sine of angle C.
Given in the question
In $\Delta \,ABC$,
$\Rightarrow \,\,{a^2} + {c^2} = 2002{b^2}$
We have to find the value of $\dfrac{{\cot A + \cot C}}{{\cot B}}$
Consider,
$\Rightarrow \,\,\dfrac{{\cot A + \cot C}}{{\cot B}}$----------(2)
By the definition of trigonometric ratios: $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$, then
$\Rightarrow \,\,\dfrac{{\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\cos C}}{{\sin C}}}}{{\dfrac{{\cos B}}{{\sin B}}}}$
By sine rule: $\sin A = \dfrac{a}{{2r}}$, $\sin B = \dfrac{b}{{2r}}$ and $\sin C = \dfrac{c}{{2r}}$
And
By cosine rule: $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$, $\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}$ and $\cos c = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$
On substituting in equation (2), we have
$\Rightarrow \,\,\dfrac{{\dfrac{{\left( {\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}} \right)}}{{\left( {\dfrac{a}{{2R}}} \right)}} + \dfrac{{\left( {\dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}} \right)}}{{\left( {\dfrac{c}{{2R}}} \right)}}}}{{\dfrac{{\left( {\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}} \right)}}{{\left( {\dfrac{b}{{2R}}} \right)}}}}$
On simplification, we get
$\Rightarrow \,\,\dfrac{{\dfrac{{{b^2} + {c^2} - {a^2}}}{{abc}} + \dfrac{{{a^2} + {b^2} - {c^2}}}{{abc}}}}{{\dfrac{{{a^2} + {c^2} - {b^2}}}{{abc}}}}$
$\Rightarrow \,\,\dfrac{{\dfrac{{{b^2} + {c^2} - {a^2} + {a^2} + {b^2} - {c^2}}}{{abc}}}}{{\dfrac{{{a^2} + {c^2} - {b^2}}}{{abc}}}}$
$\Rightarrow \,\,\dfrac{{{b^2} + {c^2} - {a^2} + {a^2} + {b^2} - {c^2}}}{{{a^2} + {c^2} - {b^2}}}$
Again, by simplification we get
$\Rightarrow \,\,\dfrac{{2{b^2}}}{{{a^2} + {c^2} - {b^2}}}$
Given ${a^2} + {c^2} = 2002{b^2}$ on substituting, we have
$\Rightarrow \,\,\dfrac{{2{b^2}}}{{2002{b^2} - {b^2}}}$
$\Rightarrow \,\,\dfrac{{2{b^2}}}{{2001{b^2}}}$
On cancelling like terms in numerator and denominator
$\Rightarrow \,\,\dfrac{2}{{2001}}$
$\,\dfrac{2}{{2001}}$ is there in the given choices.
Hence option B is the correct answer.
So, the correct answer is “Option B”.

Note : In $\Delta \,ABC$ law of sine defined as the ratio of the length of sides of a triangle to the sine of the opposite angle of a triangle. The law of sine is also known as sine rule, sine law, or sine formula.
Law of sine is used to solve $\Delta \,ABC$ is:
$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R$
Where, $\sin A = \dfrac{a}{{2R}}$, $\sin B = \dfrac{b}{{2R}}$ and $\sin C = \dfrac{c}{{2R}}$
Law of cosine defined as the square of the length of any side of a given triangle is equal to the sum of the squares of the length of the other sides minus twice the product of the other two sides multiplied by the cosine of angle included between them. Cosine rule is also called the law of cosines or Cosine Formula.
Law of cosine is used to solve $\Delta \,ABC$ is:
${a^2} = {b^2} + {c^2} - 2bc\,\cos A$
${b^2} = {a^2} + {c^2} - 2ac\,\cos B$
${c^2} = {a^2} + {b^2} - 2ab\,\cos C$
Where, $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$, $\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}$ and $\cos c = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$.