Question

If $A + B = {225^ \circ }$, then the value of $\left( {\dfrac{{\cot A}}{{1 + \cot A}}} \right)\left( {\dfrac{{\cot B}}{{1 + \cot B}}} \right)$ is equal to
(A) $1$
(B) $ - 1$
(C) $0$
(D) $\dfrac{1}{2}$

Answer 1

Hint: In the given question, we are required to find the value of trigonometric expression $\left( {\dfrac{{\cot A}}{{1 + \cot A}}} \right)\left( {\dfrac{{\cot B}}{{1 + \cot B}}} \right)$. The given problem can be solved by using the trigonometric compound formula for cotangent. We will take the cotangent function on both sides of the relation between the two angles given to us in the question.

Complete step-by-step solution:
In the given problem, we have to find value of $\left( {\dfrac{{\cot A}}{{1 + \cot A}}} \right)\left( {\dfrac{{\cot B}}{{1 + \cot B}}} \right)$.
We are given that $A + B = {225^ \circ }$, Since, the expression whose value is to be found involves the trigonometric function cotangent, so taking cotangent on both sides of the equation, we get,
$ \Rightarrow \cot \left( {A + B} \right) = \cot {225^ \circ }$
Now, using the compound angle formula for cotangent $\cot \left( {A + B} \right) = \left( {\dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}} \right)$. So, we get,
$ \Rightarrow \left( {\dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}} \right) = \cot {225^ \circ }$
We know that the value of tangent and cotangent function repeats after ${180^ \circ }$.So, we get,
$ \Rightarrow \left( {\dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}} \right) = \cot \left( {{{225}^ \circ } - {{180}^ \circ }} \right)$
\[ \Rightarrow \left( {\dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}} \right) = \cot \left( {{{45}^ \circ }} \right)\]
We know the value of $\cot \left( {{{45}^ \circ }} \right)$ as one. So, we get,
\[ \Rightarrow \left( {\dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}} \right) = 1\]
Cross multiplying the terms of equation, we get,
\[ \Rightarrow \cot A\cot B - 1 = \cot A + \cot B - - - - \left( 1 \right)\]
Now, we simplify the expression whose value is to be found. We will use the above equation later in the solution.
So, we have, $\left( {\dfrac{{\cot A}}{{1 + \cot A}}} \right)\left( {\dfrac{{\cot B}}{{1 + \cot B}}} \right)$
Multiplying the terms in numerator and denominator, we get,
$ \Rightarrow \left( {\dfrac{{\cot A\cot B}}{{1 + \cot A + \cot B + \cot A\cot B}}} \right)$
Now, substituting the value of \[\cot A + \cot B\] from equation $\left( 1 \right)$, we get,
$ \Rightarrow \left( {\dfrac{{\cot A\cot B}}{{1 + \cot A\cot B - 1 + \cot A\cot B}}} \right)$
Simplifying the expression,
$ \Rightarrow \left( {\dfrac{{\cot A\cot B}}{{2\cot A\cot B}}} \right)$
Cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow \left( {\dfrac{1}{2}} \right)$
Option (D) is the correct answer.

Note: We must have a strong grip over the concepts of trigonometry, related formulae and rules to ace these types of questions. We should know the transposition method to solve the equations in such problems. Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem.