Question

If sin α sin β − cos α cos β + 1 = 0, prove that 1 + cot α tan β = 0.

Answer 1

Given:
$$\begin{gathered} \sin \alpha s\mathrm{in}\mathrm{\beta \; -\; cos}\alpha \cos \beta \; +\; 1\; =\; 0 \\ \Rightarrow -(\cos \alpha \cos \beta -\sin \alpha \sin \beta )+1=0 \\ \Rightarrow -\cos (\alpha +\beta )+1=0 \\ \Rightarrow \cos (\alpha +\beta )=1 \\ \mathrm{Therefore},\sin (\alpha +\beta )=0....\left(1\right)(\mathrm{Since}\sin \theta =\sqrt{1-{\cos }^2\theta }) \\ \mathrm{Hence}, \\ 1+\cot \alpha \tan \beta =1+\frac{\cos \alpha \sin \beta }{\sin \alpha \cos \beta } \\ =\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\sin \alpha \cos \beta } \\ =\frac{\sin (\alpha +\beta )}{\sin \alpha \cos \beta } \\ =0...\left\{\mathrm{From}\mathrm{eq}\left(1\right)\right\} \\ \mathrm{Hence}\text{p}\mathrm{roved}. \end{gathered}$$