Question

If $A+B={225}^o$, then $\frac{\cot A}{1+\cot A}.\frac{\cot B}{1+\cot B}=$

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

Let us consider the problem:

$\cot (A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}$

Hence,

$A={225}^{\circ }-B$ simplify the above equation,

$\cot A+\cot B+1=\cot A\cot B$-------------$\left(i\right)$

Implies that

$P=\frac{\cot A\cot B}{(1+\cot A)(1+\cot B)}=\frac{\cot A\cot B}{1+\cot A+\cot B+\cot A\cot B}=\frac{\cot A\cot B}{2\cot A\cot B}=\frac{1}{2}$

Implies that

$P=\frac{\cot A\cot B}{(1+\cot A)(1+\cot B)}=\frac{\cos A\cos B}{\sin (A+B)+\cos (A-B)}$

Now,puting the values,$A={225}^{\circ }-B$

Implies that

$P=\frac{\cos {225}^{\circ }{\cos }^{2}B+\sin {225}^{\circ }\sin B\cos B}{\sin {225}^{\circ }+\cos {225}^{\circ }\cos 2B+\sin {225}^{\circ }\sin 2B}$

Therefore,

Using,$\cos {225}^{\circ }=\sin {225}^{\circ }$

Implies that

$P=\frac{\cos B(\cos B+\sin B)}{1+\cos 2B+\sin 2B}$

Implies that

$\cos 2B+1=2{\cos }^{2}B$

Hence,

$\sin 2B=2\sin B\cos B$

$P=\frac{1}{2}$