Q. If $A + B = 45{^\circ}$, then $\left( \cot{A - 1} \right)(\cot{B - 1})$ is equal to
A) 1
B) $\frac{1}{2}$
C) $- 1$
D) 2
Explanation
(d)
Given, $A + B = 45{^\circ}$
$\Rightarrow \ \cot{(A + B) = 1}$
$\Rightarrow \ \ \ \frac{\cot{A\cot{B - 1}}}{\cot{A + \cot B}} = 1$
$\Rightarrow \ \cot{Acot\ B - (\cot{A + \cot B})} = 1$ …(i)
Now, $\left( \cot{A - 1} \right)\left( \cot{B - 1} \right) = \cot A\cot{B - \left( \cot{A + \cot B} \right) + 1}$
$= 1 + 1 = 2$ [from Eq. (i)]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.