To find the value of $$\frac{\cot{A}\cot{B}}{(1+\cot{A})(1+\cot{B})}$$, given $A+B=225^\circ$
i write the above fraction as after converted to sin and cos as $$\frac{\cos{A}\cos{B}}{sin(A+B)cos(A-B)}$$
Now since $A+B=225^\circ$. I can find $\sin(A+B)$ , but how do i evaluate other terms
Thanks a ton!
We know that $$\cot (A+B) = \frac {\cot A \cot B -1 }{\cot A + \cot B} $$
As $A+B =225^\circ $, simplifying the above equation gives us $$\cot A + \cot B +1 =\cot A \cot B \tag {1} $$
Now from the original expression, we get, $$P = \frac {\cot A \cot B}{(1+\cot A)(1+\cot B)} = \frac {\cot A \cot B}{1+\cot A +\cot B +\cot A \cot B} = \frac {\cot A\cot B}{2\cot A \cot B} = \frac {1}{2} $$
If you want to continue with your method, we have, $$P = \frac {\cot A \cot B}{(1+\cot A)(1+\cot B)} = \frac {\cos A \cos B}{\sin (A+B) + \cos (A-B)}$$ Now substituting $A=225^\circ -B $, we get, $$P = \frac {\cos 225^\circ \cos^2 B + \sin 225^\circ \sin B \cos B}{\sin 225^\circ + \cos 225^\circ \cos 2B + \sin 225^\circ \sin 2B } $$ As $\cos 225^\circ = \sin 225^\circ $, we have, $$P =\frac {\cos B (\cos B + \sin B)}{1 + \cos 2B + \sin 2B} $$ and using $\cos 2B +1 =2\cos^2 B $ and $\sin 2B =2\sin B \cos B $ gives us the result as $$\boxed{P = \frac {1}{2}} $$ the same as before.
Hope it helps.
$$A+B =225^\circ=\pi+\dfrac{\pi}{4} \to \\ \tan(A+B)=\dfrac{\tan(A)+\tan(B)}{1-\tan(A).\tan(B)}=\tan(\pi+\dfrac{\pi}{4})=1 \\ \to \\\tan(A)+\tan(B)=1-\tan(A).\tan(B) \\(***)\tan(A)+\tan(B)+\tan(A).\tan(B)=1$$ $$\frac{\cot{A}\cot{B}}{(1+\cot{A})(1+\cot{B})}=\\\frac{\cot{A}\cot{B}}{(1+\cot{A})(1+\cot{B})}.\dfrac{\tan A}{\tan A}.\dfrac{\tan B}{\tan B}=\\ \dfrac{1}{(\tan A+1)(\tan B+1)}= \\\dfrac{1}{tan(A)+\tan(B)+\tan(A).\tan(B)+1}$$now from (***) $$=\dfrac{1}{tan(A)+\tan(B)+\tan(A).\tan(B)+1}=\\\dfrac{1}{1+1}=\dfrac12$$
$$\cot(180^\circ+45^\circ)=\cot(A+B)=\dfrac{\cot A\cot B-1}{\cot B+\cot A}$$
$$\implies\cot B+\cot A=\cot A\cot B-1$$
Now find $(1+\cot A)(1+\cot B)$