Question

If $\sin \alpha \sin \beta -\cos \alpha \cos \beta +1=0$, then $\cot \alpha \tan \beta =-a$
Find a

Answer 1

Given, $\sin \alpha \sin \beta -\cos \alpha \cos \beta +1=0$
or $\cos \alpha \cos \beta -\sin \alpha \sin \beta =1$
or $\cos (\alpha +\beta )=1$ (i)
Now $1+\cot \alpha \tan \beta =1+\frac{\cos \alpha }{\sin \alpha }\times \frac{\sin \beta }{\cos \beta }$
$=\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\sin \alpha \cos \beta }$
$=\frac{\sin (\alpha +\beta )}{\sin \alpha \cos \beta }$
$=\frac{0}{\sin \alpha \cos \beta }=0$ $[\because {\sin }^2(\alpha +\beta )=1-{\cos }^2(\alpha +\beta )=1-1=0]$
Therefore. $a=1$
Ans: 1