Question

If $\sin \beta -\cos \alpha \cos \beta +1=0$ then show that $1+\cot \alpha \tan \beta =0$.

Answer 1

$\begin{array}{c}\underset{–}{wehavegivenquestion:}\\ sin\alpha sin\beta -cos\alpha cos\beta +1=0\\ \Rightarrow sin\alpha sin\beta -cos\alpha cos\beta =-1\\ \Rightarrow cos\alpha cos\beta -sin\alpha sin\beta =1\\ Now,\\ \Rightarrow \cos (\alpha +\beta )=1|\begin{array}{c}formformula:\cos (A+B)=\cos A\cos B-\sin A\sin B\\ weknow:\cos 0=1\end{array}\\ \Rightarrow \alpha +\beta =0\\ \therefore \beta =-\alpha \\ NowfromL.H.S:-\\ \Rightarrow 1+\cot \alpha \tan \beta =0\\ \Rightarrow 1+\cot \alpha \tan (-\alpha )=0\\ \Rightarrow 1-\cot \alpha \tan \alpha =0\\ \Rightarrow 1-\frac{1}{\tan \alpha }\tan \alpha =0\\ \therefore 1-1=0R.H.S\end{array}$