Question

If \[A + B = {225^ \circ },\] then \[\dfrac{{\cot A}}{{1 + \cot A}} \cdot \dfrac{{\cot B}}{{1 + \cot B}} = \]
A. $1$
B. $ - 1$
C. 0
D. $\dfrac{1}{2}$

Answer 1

Hint: In order to solve this type of question, first we will consider the given question. Then, we will simplify it. Next, we will consider the given equation and simplify it by taking tan on both sides. Now, we will apply trigonometric identity for compound angles, substitute the values in it and simplify it to get the correct answer.

Formula used:
$\left[ {\because \cot A = \dfrac{1}{{\tan A}}} \right]$
$\left[ {\because \tan \left( {{{180}^ \circ } + \theta } \right) = \tan \theta } \right]$
$\left[ {\because \tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}} \right]$
$\left[ {\because \tan {{45}^ \circ } = 1} \right]$

Complete step by step solution:
Consider,
\[\dfrac{{\cot A}}{{1 + \cot A}} \cdot \dfrac{{\cot B}}{{1 + \cot B}}\]
$ = \left( {\dfrac{{\dfrac{1}{{\tan A}}}}{{1 + \dfrac{1}{{\tan A}}}}} \right) \cdot \left( {\dfrac{{\dfrac{1}{{\tan B}}}}{{1 + \dfrac{1}{{\tan B}}}}} \right)$ $\left[ {\because \cot A = \dfrac{1}{{\tan A}}} \right]$
Solving it,
$ = \dfrac{1}{{\left( {1 + \tan A} \right)\left( {1 + \tan B} \right)}}$ ………………..equation$\left( 1 \right)$
We are given,
\[A + B = {225^ \circ }\]
\[B = {225^ \circ } - A\]
Taking $\tan $ on both sides,
\[\tan B = \tan \left( {{{225}^ \circ } - A} \right)\]
\[\tan B = \tan \left( {{{180}^ \circ } + {{45}^ \circ } - A} \right)\]
\[\tan B = \tan \left( {{{45}^ \circ } - A} \right)\] $\left[ {\because \tan \left( {{{180}^ \circ } + \theta } \right) = \tan \theta } \right]$
Applying trigonometric identity,
$\tan B = \dfrac{{\tan {{45}^ \circ } - \tan A}}{{1 + \tan {{45}^ \circ }\tan A}}$ $\left[ {\because \tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}} \right]$
$\tan B = \dfrac{{1 - \tan A}}{{1 + \tan A}}$ $\left[ {\because \tan {{45}^ \circ } = 1} \right]$
Substituting this value in equation $\left( 1 \right)$,
\[\dfrac{{\cot A}}{{1 + \cot A}} \cdot \dfrac{{\cot B}}{{1 + \cot B}} = \dfrac{1}{{\left( {1 + \tan A} \right)\left( {1 + \left( {\dfrac{{1 - \tan A}}{{1 + \tan A}}} \right)} \right)}}\]
\[ = \dfrac{{\left( {1 + \tan A} \right)}}{{\left( {1 + \tan A} \right)\left( {1 + \tan A + 1 - \tan A} \right)}}\]
On simplifying it,
\[\dfrac{{\cot A}}{{1 + \cot A}} \cdot \dfrac{{\cot B}}{{1 + \cot B}} = \dfrac{1}{2}\]
$\therefore $ The correct option is D.

Note: We can also solve this question by taking cotangent on both sides of the given equation \[A+B={{225}^{0}}\] to find the value of \[\frac{\cot A}{1+\cot A}\times \frac{\cot B}{1+\cot B}\].
\[\cot \left( A+B \right)=\cot {{225}^{0}}\]
Now we know that \[\cot \left( A+B \right)=\frac{\cot A\cot B-1}{\cot A+\cot B}\] and \[\cot {{225}^{0}}=1\] so we will use these in the above equation.
\[\begin{align}
  & \cot \left( A+B \right)=\cot {{225}^{0}} \\
 & \frac{\cot A\cot B-1}{\cot A+\cot B}=1 \\
 & \cot A\cot B-1=\cot A+\cot B \\
 & \cot A\cot B=1+\cot A+\cot B.....(i)
\end{align}\]
We will now take \[\frac{\cot A}{1+\cot A}\times \frac{\cot B}{1+\cot B}\] and find its product.
\[\begin{align}
  & =\frac{\cot A}{1+\cot A}\times \frac{\cot B}{1+\cot B} \\
 & =\frac{\cot A\cot B}{1+\cot B+\cot A+\cot A\cot B} \\
\end{align}\]
Substituting equation (i) in the above equation,
\[\begin{align}
  & =\frac{\cot A\cot B}{1+\cot B+\cot A+\cot A\cot B} \\
 & =\frac{\cot A\cot B}{\cot A\cot B+\cot A\cot B} \\
 & =\frac{\cot A\cot B}{2\cot A\cot B} \\
 & =\frac{1}{2} \\
\end{align}\]