Question

If sin $\alpha sin\beta cos\beta +1=0$, prove that 1 + cot $\alpha tan\beta =0$.

Answer 1

We have,
$sin\alpha sin\beta -cos\alpha cos\beta +1=0\Rightarrow -(cos\alpha cos\beta -sin\beta )=-1\Rightarrow cos(\alpha +\beta )=1\therefore sin(\alpha +\beta )=\sqrt{1-co{s}^2(\alpha +\beta )}=\sqrt{1-1^2}=0\Rightarrow sin(\alpha +\beta )=0$
Now,
$1+cot\alpha tan\beta =1+\frac{cos\alpha }{sin\alpha }\times \frac{sin\beta }{cos\beta }=\frac{sin\alpha \times cos\beta +cos\alpha \times sinbeta}{sin\alpha \times cos\beta }$
= $\frac{sin(\alpha +\beta )}{sin\alpha \times cos\beta }$
= $\frac{0}{sin\alpha \times cos\beta }$ [Using equatioon (ii)]
= 0
$\therefore 1+cot\alpha tan\beta =0$ Hence proved.